Is it possible to perform a computation to know if an object is at rest and its location given a force versus...












1














The problem is as follows:




The force graphed in Figure 1. is applied to a $textrm{2.0-kg}$
block that was sliding to the right (the $textrm{+-direction}$) over
a frictionless surface with speed $5.0,frac{m}{s}$ at $x=0$ (a) Is
the block ever at rest? If so,where? (b) Find a position (other than
$x=0$) when the block is again moving to the right at
$5.0,frac{m}{s}$.




Sketch of the problem



The figure alluded in the problem is shown above.



Now the thing with this problem is I really don't know where to begin or if there is something missing in my maths.



Hence my question is how do I know that the block is at rest?.



My first guess is that this happens when the net force on the block is $0$ or that it is not carrying (unsure if this term is right) kinetic energy?. However can this happen in a frictionless surface?. My other question arises from the fact, how do I understand the negative force in the diagram?. Does it mean that is a force that is being applied to the left side of the block?.



So far the only thing I could come up with wad to use Newton's second law:



$f=ma$



Since what it is being asked is the position, I thought to use:



$v^{2}_f=v^{2}_{0}+2aDelta x$



For $v^{2}_f=0$ and $v_{0}=5,frac{m}{s}$,



$0=left(5right)^{2}_{0}+2aDelta x$



But this would mean to use a, from where? From the diagram?.



If so, at $-40 N$ :



$a=frac{-40}{2}=-20,frac{m}{s^2}$



Then:



$0=left(5right)^{2}+2left(-20right)Delta x$



Hence:



$Delta x=frac{25}{40}=0.625$



Which would translate as a position $x=0$



$x_{1}-x_{0}=0.625-0=0.625,m$



So this would be at the right from the starting point at $x=0$ and although this seems consistent with what it is alluded in the diagram (the "domain" is in the positive numbers) it doesn't check with the answers in my book.



According to it, the answer for part (a) is $1,textrm{m}$ and for part (b) is $2.917,textrm{m}$. But again, none of this seems close to what I obtained. What could I be missing in my calculation?.



Anyways. Am I understanding this mathematically or physically wrong?. Can somebody help me to clear these doubts?.










share|cite|improve this question
























  • "the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
    – harshit54
    2 days ago












  • @harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
    – Chris Steinbeck Bell
    2 days ago










  • No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
    – harshit54
    2 days ago










  • In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
    – harshit54
    2 days ago












  • @harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
    – Chris Steinbeck Bell
    2 days ago
















1














The problem is as follows:




The force graphed in Figure 1. is applied to a $textrm{2.0-kg}$
block that was sliding to the right (the $textrm{+-direction}$) over
a frictionless surface with speed $5.0,frac{m}{s}$ at $x=0$ (a) Is
the block ever at rest? If so,where? (b) Find a position (other than
$x=0$) when the block is again moving to the right at
$5.0,frac{m}{s}$.




Sketch of the problem



The figure alluded in the problem is shown above.



Now the thing with this problem is I really don't know where to begin or if there is something missing in my maths.



Hence my question is how do I know that the block is at rest?.



My first guess is that this happens when the net force on the block is $0$ or that it is not carrying (unsure if this term is right) kinetic energy?. However can this happen in a frictionless surface?. My other question arises from the fact, how do I understand the negative force in the diagram?. Does it mean that is a force that is being applied to the left side of the block?.



So far the only thing I could come up with wad to use Newton's second law:



$f=ma$



Since what it is being asked is the position, I thought to use:



$v^{2}_f=v^{2}_{0}+2aDelta x$



For $v^{2}_f=0$ and $v_{0}=5,frac{m}{s}$,



$0=left(5right)^{2}_{0}+2aDelta x$



But this would mean to use a, from where? From the diagram?.



If so, at $-40 N$ :



$a=frac{-40}{2}=-20,frac{m}{s^2}$



Then:



$0=left(5right)^{2}+2left(-20right)Delta x$



Hence:



$Delta x=frac{25}{40}=0.625$



Which would translate as a position $x=0$



$x_{1}-x_{0}=0.625-0=0.625,m$



So this would be at the right from the starting point at $x=0$ and although this seems consistent with what it is alluded in the diagram (the "domain" is in the positive numbers) it doesn't check with the answers in my book.



According to it, the answer for part (a) is $1,textrm{m}$ and for part (b) is $2.917,textrm{m}$. But again, none of this seems close to what I obtained. What could I be missing in my calculation?.



Anyways. Am I understanding this mathematically or physically wrong?. Can somebody help me to clear these doubts?.










share|cite|improve this question
























  • "the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
    – harshit54
    2 days ago












  • @harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
    – Chris Steinbeck Bell
    2 days ago










  • No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
    – harshit54
    2 days ago










  • In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
    – harshit54
    2 days ago












  • @harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
    – Chris Steinbeck Bell
    2 days ago














1












1








1







The problem is as follows:




The force graphed in Figure 1. is applied to a $textrm{2.0-kg}$
block that was sliding to the right (the $textrm{+-direction}$) over
a frictionless surface with speed $5.0,frac{m}{s}$ at $x=0$ (a) Is
the block ever at rest? If so,where? (b) Find a position (other than
$x=0$) when the block is again moving to the right at
$5.0,frac{m}{s}$.




Sketch of the problem



The figure alluded in the problem is shown above.



Now the thing with this problem is I really don't know where to begin or if there is something missing in my maths.



Hence my question is how do I know that the block is at rest?.



My first guess is that this happens when the net force on the block is $0$ or that it is not carrying (unsure if this term is right) kinetic energy?. However can this happen in a frictionless surface?. My other question arises from the fact, how do I understand the negative force in the diagram?. Does it mean that is a force that is being applied to the left side of the block?.



So far the only thing I could come up with wad to use Newton's second law:



$f=ma$



Since what it is being asked is the position, I thought to use:



$v^{2}_f=v^{2}_{0}+2aDelta x$



For $v^{2}_f=0$ and $v_{0}=5,frac{m}{s}$,



$0=left(5right)^{2}_{0}+2aDelta x$



But this would mean to use a, from where? From the diagram?.



If so, at $-40 N$ :



$a=frac{-40}{2}=-20,frac{m}{s^2}$



Then:



$0=left(5right)^{2}+2left(-20right)Delta x$



Hence:



$Delta x=frac{25}{40}=0.625$



Which would translate as a position $x=0$



$x_{1}-x_{0}=0.625-0=0.625,m$



So this would be at the right from the starting point at $x=0$ and although this seems consistent with what it is alluded in the diagram (the "domain" is in the positive numbers) it doesn't check with the answers in my book.



According to it, the answer for part (a) is $1,textrm{m}$ and for part (b) is $2.917,textrm{m}$. But again, none of this seems close to what I obtained. What could I be missing in my calculation?.



Anyways. Am I understanding this mathematically or physically wrong?. Can somebody help me to clear these doubts?.










share|cite|improve this question















The problem is as follows:




The force graphed in Figure 1. is applied to a $textrm{2.0-kg}$
block that was sliding to the right (the $textrm{+-direction}$) over
a frictionless surface with speed $5.0,frac{m}{s}$ at $x=0$ (a) Is
the block ever at rest? If so,where? (b) Find a position (other than
$x=0$) when the block is again moving to the right at
$5.0,frac{m}{s}$.




Sketch of the problem



The figure alluded in the problem is shown above.



Now the thing with this problem is I really don't know where to begin or if there is something missing in my maths.



Hence my question is how do I know that the block is at rest?.



My first guess is that this happens when the net force on the block is $0$ or that it is not carrying (unsure if this term is right) kinetic energy?. However can this happen in a frictionless surface?. My other question arises from the fact, how do I understand the negative force in the diagram?. Does it mean that is a force that is being applied to the left side of the block?.



So far the only thing I could come up with wad to use Newton's second law:



$f=ma$



Since what it is being asked is the position, I thought to use:



$v^{2}_f=v^{2}_{0}+2aDelta x$



For $v^{2}_f=0$ and $v_{0}=5,frac{m}{s}$,



$0=left(5right)^{2}_{0}+2aDelta x$



But this would mean to use a, from where? From the diagram?.



If so, at $-40 N$ :



$a=frac{-40}{2}=-20,frac{m}{s^2}$



Then:



$0=left(5right)^{2}+2left(-20right)Delta x$



Hence:



$Delta x=frac{25}{40}=0.625$



Which would translate as a position $x=0$



$x_{1}-x_{0}=0.625-0=0.625,m$



So this would be at the right from the starting point at $x=0$ and although this seems consistent with what it is alluded in the diagram (the "domain" is in the positive numbers) it doesn't check with the answers in my book.



According to it, the answer for part (a) is $1,textrm{m}$ and for part (b) is $2.917,textrm{m}$. But again, none of this seems close to what I obtained. What could I be missing in my calculation?.



Anyways. Am I understanding this mathematically or physically wrong?. Can somebody help me to clear these doubts?.







algebra-precalculus physics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked 2 days ago









Chris Steinbeck Bell

725315




725315












  • "the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
    – harshit54
    2 days ago












  • @harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
    – Chris Steinbeck Bell
    2 days ago










  • No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
    – harshit54
    2 days ago










  • In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
    – harshit54
    2 days ago












  • @harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
    – Chris Steinbeck Bell
    2 days ago


















  • "the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
    – harshit54
    2 days ago












  • @harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
    – Chris Steinbeck Bell
    2 days ago










  • No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
    – harshit54
    2 days ago










  • In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
    – harshit54
    2 days ago












  • @harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
    – Chris Steinbeck Bell
    2 days ago
















"the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
– harshit54
2 days ago






"the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
– harshit54
2 days ago














@harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
– Chris Steinbeck Bell
2 days ago




@harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
– Chris Steinbeck Bell
2 days ago












No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
– harshit54
2 days ago




No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
– harshit54
2 days ago












In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
– harshit54
2 days ago






In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
– harshit54
2 days ago














@harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
– Chris Steinbeck Bell
2 days ago




@harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
– Chris Steinbeck Bell
2 days ago










2 Answers
2






active

oldest

votes


















1















$v^{2}_f=v^{2}_{0}+2aDelta x$




That applies when you have constant acceleration. The graph shows that force, and therefore acceleration, is changing. So you can't use this formula.



You have that the horizontal axis (width) is distance, and the vertical (height) is force. Height times width is area, and force times distance is work. So area in this diagram represents work (keep in mind that area below the x-axis is negative work). In general, when you have a diagram where the horizontal has units $a$, and the vertical has units $b$, then area represents something with units $ab$.



You're given that at $x=0$, the velocity is +5. From that, and the mass, you can figure out the initial kinetic energy. For the object to be at rest, the new kinetic energy must be 0, which means that work so far must cancel out the initial kinetic energy. So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy.




how do I understand the negative force in the diagram?




A negative force means a force in the "negative" direction. What direction is considered "negative" is somewhat arbitrary, and can be chosen according to what's convenient for the problem. If it's not stated, it's generally assumed that right is positive and left is negative. And indeed, the problem says "the right (the +-direction)", so that tells you that right is the positive direction, and left is the negative direction.






share|cite|improve this answer





















  • @Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
    – Acccumulation
    2 days ago










  • @Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
    – Chris Steinbeck Bell
    2 days ago










  • @Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
    – amd
    2 days ago



















0














Hint.



You have



$$
m ddot x = f(x) = begin{cases}
5 (6 x-8) & x le 2 \
20 & 2lt x le 3 \
10 (2 x-4) & x gt 3
end{cases}
$$



then



$$
mdot x ddot x = f(x)dot xto frac m2 frac{d}{dt}dot x ^2 = frac{d}{dt}F(x)
$$



then integrating



$$
frac 12 m (v^2-v_0^2) = F(x) = begin{cases}
15 x^2-40 x & xleq 2 \
20 x-60 & 2<xleq 3 \
20 left(frac{x^2}{2}-2 xright)+30 & 3 < x
end{cases}
$$



Attached the plot for $f(x)$ in blue, the plot for $F(x)$ in red and the plot for $v(x)$ in black showing the points in which $v = v_0$ which occurs at $x = 0$ and $x = 3$



enter image description hereenter image description hereenter image description here






share|cite|improve this answer























  • I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
    – Chris Steinbeck Bell
    2 days ago










  • I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell I hope now helps more.
    – Cesareo
    2 days ago










  • @ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
    – Cesareo
    2 days ago






  • 1




    @ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
    – Andrei
    yesterday











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060861%2fis-it-possible-to-perform-a-computation-to-know-if-an-object-is-at-rest-and-its%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1















$v^{2}_f=v^{2}_{0}+2aDelta x$




That applies when you have constant acceleration. The graph shows that force, and therefore acceleration, is changing. So you can't use this formula.



You have that the horizontal axis (width) is distance, and the vertical (height) is force. Height times width is area, and force times distance is work. So area in this diagram represents work (keep in mind that area below the x-axis is negative work). In general, when you have a diagram where the horizontal has units $a$, and the vertical has units $b$, then area represents something with units $ab$.



You're given that at $x=0$, the velocity is +5. From that, and the mass, you can figure out the initial kinetic energy. For the object to be at rest, the new kinetic energy must be 0, which means that work so far must cancel out the initial kinetic energy. So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy.




how do I understand the negative force in the diagram?




A negative force means a force in the "negative" direction. What direction is considered "negative" is somewhat arbitrary, and can be chosen according to what's convenient for the problem. If it's not stated, it's generally assumed that right is positive and left is negative. And indeed, the problem says "the right (the +-direction)", so that tells you that right is the positive direction, and left is the negative direction.






share|cite|improve this answer





















  • @Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
    – Acccumulation
    2 days ago










  • @Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
    – Chris Steinbeck Bell
    2 days ago










  • @Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
    – amd
    2 days ago
















1















$v^{2}_f=v^{2}_{0}+2aDelta x$




That applies when you have constant acceleration. The graph shows that force, and therefore acceleration, is changing. So you can't use this formula.



You have that the horizontal axis (width) is distance, and the vertical (height) is force. Height times width is area, and force times distance is work. So area in this diagram represents work (keep in mind that area below the x-axis is negative work). In general, when you have a diagram where the horizontal has units $a$, and the vertical has units $b$, then area represents something with units $ab$.



You're given that at $x=0$, the velocity is +5. From that, and the mass, you can figure out the initial kinetic energy. For the object to be at rest, the new kinetic energy must be 0, which means that work so far must cancel out the initial kinetic energy. So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy.




how do I understand the negative force in the diagram?




A negative force means a force in the "negative" direction. What direction is considered "negative" is somewhat arbitrary, and can be chosen according to what's convenient for the problem. If it's not stated, it's generally assumed that right is positive and left is negative. And indeed, the problem says "the right (the +-direction)", so that tells you that right is the positive direction, and left is the negative direction.






share|cite|improve this answer





















  • @Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
    – Acccumulation
    2 days ago










  • @Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
    – Chris Steinbeck Bell
    2 days ago










  • @Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
    – amd
    2 days ago














1












1








1







$v^{2}_f=v^{2}_{0}+2aDelta x$




That applies when you have constant acceleration. The graph shows that force, and therefore acceleration, is changing. So you can't use this formula.



You have that the horizontal axis (width) is distance, and the vertical (height) is force. Height times width is area, and force times distance is work. So area in this diagram represents work (keep in mind that area below the x-axis is negative work). In general, when you have a diagram where the horizontal has units $a$, and the vertical has units $b$, then area represents something with units $ab$.



You're given that at $x=0$, the velocity is +5. From that, and the mass, you can figure out the initial kinetic energy. For the object to be at rest, the new kinetic energy must be 0, which means that work so far must cancel out the initial kinetic energy. So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy.




how do I understand the negative force in the diagram?




A negative force means a force in the "negative" direction. What direction is considered "negative" is somewhat arbitrary, and can be chosen according to what's convenient for the problem. If it's not stated, it's generally assumed that right is positive and left is negative. And indeed, the problem says "the right (the +-direction)", so that tells you that right is the positive direction, and left is the negative direction.






share|cite|improve this answer













$v^{2}_f=v^{2}_{0}+2aDelta x$




That applies when you have constant acceleration. The graph shows that force, and therefore acceleration, is changing. So you can't use this formula.



You have that the horizontal axis (width) is distance, and the vertical (height) is force. Height times width is area, and force times distance is work. So area in this diagram represents work (keep in mind that area below the x-axis is negative work). In general, when you have a diagram where the horizontal has units $a$, and the vertical has units $b$, then area represents something with units $ab$.



You're given that at $x=0$, the velocity is +5. From that, and the mass, you can figure out the initial kinetic energy. For the object to be at rest, the new kinetic energy must be 0, which means that work so far must cancel out the initial kinetic energy. So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy.




how do I understand the negative force in the diagram?




A negative force means a force in the "negative" direction. What direction is considered "negative" is somewhat arbitrary, and can be chosen according to what's convenient for the problem. If it's not stated, it's generally assumed that right is positive and left is negative. And indeed, the problem says "the right (the +-direction)", so that tells you that right is the positive direction, and left is the negative direction.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









Acccumulation

6,8062618




6,8062618












  • @Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
    – Acccumulation
    2 days ago










  • @Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
    – Chris Steinbeck Bell
    2 days ago










  • @Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
    – amd
    2 days ago


















  • @Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
    – Acccumulation
    2 days ago










  • @Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
    – Chris Steinbeck Bell
    2 days ago










  • @Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
    – amd
    2 days ago
















@Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
– Chris Steinbeck Bell
2 days ago




@Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
– Chris Steinbeck Bell
2 days ago












@ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
– Acccumulation
2 days ago




@ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
– Acccumulation
2 days ago












@Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
– Chris Steinbeck Bell
2 days ago




@Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
– Chris Steinbeck Bell
2 days ago












@Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
– Chris Steinbeck Bell
2 days ago




@Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
– Chris Steinbeck Bell
2 days ago












@ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
– amd
2 days ago




@ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
– amd
2 days ago











0














Hint.



You have



$$
m ddot x = f(x) = begin{cases}
5 (6 x-8) & x le 2 \
20 & 2lt x le 3 \
10 (2 x-4) & x gt 3
end{cases}
$$



then



$$
mdot x ddot x = f(x)dot xto frac m2 frac{d}{dt}dot x ^2 = frac{d}{dt}F(x)
$$



then integrating



$$
frac 12 m (v^2-v_0^2) = F(x) = begin{cases}
15 x^2-40 x & xleq 2 \
20 x-60 & 2<xleq 3 \
20 left(frac{x^2}{2}-2 xright)+30 & 3 < x
end{cases}
$$



Attached the plot for $f(x)$ in blue, the plot for $F(x)$ in red and the plot for $v(x)$ in black showing the points in which $v = v_0$ which occurs at $x = 0$ and $x = 3$



enter image description hereenter image description hereenter image description here






share|cite|improve this answer























  • I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
    – Chris Steinbeck Bell
    2 days ago










  • I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell I hope now helps more.
    – Cesareo
    2 days ago










  • @ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
    – Cesareo
    2 days ago






  • 1




    @ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
    – Andrei
    yesterday
















0














Hint.



You have



$$
m ddot x = f(x) = begin{cases}
5 (6 x-8) & x le 2 \
20 & 2lt x le 3 \
10 (2 x-4) & x gt 3
end{cases}
$$



then



$$
mdot x ddot x = f(x)dot xto frac m2 frac{d}{dt}dot x ^2 = frac{d}{dt}F(x)
$$



then integrating



$$
frac 12 m (v^2-v_0^2) = F(x) = begin{cases}
15 x^2-40 x & xleq 2 \
20 x-60 & 2<xleq 3 \
20 left(frac{x^2}{2}-2 xright)+30 & 3 < x
end{cases}
$$



Attached the plot for $f(x)$ in blue, the plot for $F(x)$ in red and the plot for $v(x)$ in black showing the points in which $v = v_0$ which occurs at $x = 0$ and $x = 3$



enter image description hereenter image description hereenter image description here






share|cite|improve this answer























  • I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
    – Chris Steinbeck Bell
    2 days ago










  • I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell I hope now helps more.
    – Cesareo
    2 days ago










  • @ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
    – Cesareo
    2 days ago






  • 1




    @ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
    – Andrei
    yesterday














0












0








0






Hint.



You have



$$
m ddot x = f(x) = begin{cases}
5 (6 x-8) & x le 2 \
20 & 2lt x le 3 \
10 (2 x-4) & x gt 3
end{cases}
$$



then



$$
mdot x ddot x = f(x)dot xto frac m2 frac{d}{dt}dot x ^2 = frac{d}{dt}F(x)
$$



then integrating



$$
frac 12 m (v^2-v_0^2) = F(x) = begin{cases}
15 x^2-40 x & xleq 2 \
20 x-60 & 2<xleq 3 \
20 left(frac{x^2}{2}-2 xright)+30 & 3 < x
end{cases}
$$



Attached the plot for $f(x)$ in blue, the plot for $F(x)$ in red and the plot for $v(x)$ in black showing the points in which $v = v_0$ which occurs at $x = 0$ and $x = 3$



enter image description hereenter image description hereenter image description here






share|cite|improve this answer














Hint.



You have



$$
m ddot x = f(x) = begin{cases}
5 (6 x-8) & x le 2 \
20 & 2lt x le 3 \
10 (2 x-4) & x gt 3
end{cases}
$$



then



$$
mdot x ddot x = f(x)dot xto frac m2 frac{d}{dt}dot x ^2 = frac{d}{dt}F(x)
$$



then integrating



$$
frac 12 m (v^2-v_0^2) = F(x) = begin{cases}
15 x^2-40 x & xleq 2 \
20 x-60 & 2<xleq 3 \
20 left(frac{x^2}{2}-2 xright)+30 & 3 < x
end{cases}
$$



Attached the plot for $f(x)$ in blue, the plot for $F(x)$ in red and the plot for $v(x)$ in black showing the points in which $v = v_0$ which occurs at $x = 0$ and $x = 3$



enter image description hereenter image description hereenter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









Cesareo

8,3493516




8,3493516












  • I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
    – Chris Steinbeck Bell
    2 days ago










  • I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell I hope now helps more.
    – Cesareo
    2 days ago










  • @ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
    – Cesareo
    2 days ago






  • 1




    @ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
    – Andrei
    yesterday


















  • I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
    – Chris Steinbeck Bell
    2 days ago










  • I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
    – Chris Steinbeck Bell
    2 days ago










  • @ChrisSteinbeckBell I hope now helps more.
    – Cesareo
    2 days ago










  • @ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
    – Cesareo
    2 days ago






  • 1




    @ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
    – Andrei
    yesterday
















I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
– Chris Steinbeck Bell
2 days ago




I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
– Chris Steinbeck Bell
2 days ago












I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
– Chris Steinbeck Bell
2 days ago




I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
– Chris Steinbeck Bell
2 days ago












@ChrisSteinbeckBell I hope now helps more.
– Cesareo
2 days ago




@ChrisSteinbeckBell I hope now helps more.
– Cesareo
2 days ago












@ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
– Cesareo
2 days ago




@ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
– Cesareo
2 days ago




1




1




@ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
– Andrei
yesterday




@ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
– Andrei
yesterday


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060861%2fis-it-possible-to-perform-a-computation-to-know-if-an-object-is-at-rest-and-its%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

1300-talet

1300-talet

Display a custom attribute below product name in the front-end Magento 1.9.3.8