Further Mechanics Conservation of Energy Question
Been struggling with the second part of a mechanical question. It goes:
Particle A of mass 2m can move on the rough surface of a plane inclined at angle theta to the horizontal, where sin(theta) is 3/5. A second particle B, of mass 5m, hangs freely attached to a light smooth pulley fixed at D. The other end of the string is attached to A.
The coefficient of friction between A and the plane is 3/8. Particle B is 2 metres from the ground, A is 4 metres from D.
When the system is released from rest with the string taught, A moves up the plane.
The first question is what's A's initial acceleration. I got that fine, as 4.48, which is right.
The second question is:
When B has descended 1m the string breaks. Using the principle of conservation of energy, find the distance A moves before coming to rest.
The correct answer for that is 1.51 metres, to two s.f. I just have no idea how to get there.
My first thoughts are that I need to find A's potential energy and kinetic energy when the rope snaps, but suvat doesn't seem to work to get it's final speed. Then when I use friction to see how far it moves it doesn't return the right answer.
Obviously I'm going about it the wrong way, so any tips and help would be very appreciated!
Particle A of mass 2m can move on the rough surface of a plane inclined at angle theta to the horizontal, where sin(theta) is 3/5. A second particle B, of mass 5m, hangs freely attached to a light smooth pulley fixed at D. The other end of the string is attached to A.
The coefficient of friction between A and the plane is 3/8. Particle B is 2 metres from the ground, A is 4 metres from D.
When the system is released from rest with the string taught, A moves up the plane.
The first question is what's A's initial acceleration. I got that fine, as 4.48, which is right.
The second question is:
When B has descended 1m the string breaks. Using the principle of conservation of energy, find the distance A moves before coming to rest.
The correct answer for that is 1.51 metres, to two s.f. I just have no idea how to get there.
My first thoughts are that I need to find A's potential energy and kinetic energy when the rope snaps, but suvat doesn't seem to work to get it's final speed. Then when I use friction to see how far it moves it doesn't return the right answer.
Obviously I'm going about it the wrong way, so any tips and help would be very appreciated!
Reply 1
21
Original post
by mossyrevelationsBeen struggling with the second part of a mechanical question. It goes:
Particle A of mass 2m can move on the rough surface of a plane inclined at angle theta to the horizontal, where sin(theta) is 3/5. A second particle B, of mass 5m, hangs freely attached to a light smooth pulley fixed at D. The other end of the string is attached to A.
The coefficient of friction between A and the plane is 3/8. Particle B is 2 metres from the ground, A is 4 metres from D.
When the system is released from rest with the string taught, A moves up the plane.
The first question is what's A's initial acceleration. I got that fine, as 4.48, which is right.
The second question is:
When B has descended 1m the string breaks. Using the principle of conservation of energy, find the distance A moves before coming to rest.
The correct answer for that is 1.51 metres, to two s.f. I just have no idea how to get there.
My first thoughts are that I need to find A's potential energy and kinetic energy when the rope snaps, but suvat doesn't seem to work to get it's final speed. Then when I use friction to see how far it moves it doesn't return the right answer.
Obviously I'm going about it the wrong way, so any tips and help would be very appreciated!
Particle A of mass 2m can move on the rough surface of a plane inclined at angle theta to the horizontal, where sin(theta) is 3/5. A second particle B, of mass 5m, hangs freely attached to a light smooth pulley fixed at D. The other end of the string is attached to A.
The coefficient of friction between A and the plane is 3/8. Particle B is 2 metres from the ground, A is 4 metres from D.
When the system is released from rest with the string taught, A moves up the plane.
The first question is what's A's initial acceleration. I got that fine, as 4.48, which is right.
The second question is:
When B has descended 1m the string breaks. Using the principle of conservation of energy, find the distance A moves before coming to rest.
The correct answer for that is 1.51 metres, to two s.f. I just have no idea how to get there.
My first thoughts are that I need to find A's potential energy and kinetic energy when the rope snaps, but suvat doesn't seem to work to get it's final speed. Then when I use friction to see how far it moves it doesn't return the right answer.
Obviously I'm going about it the wrong way, so any tips and help would be very appreciated!
It should involve be KE, GPE and Work (done against friction).
Initially (rope snaps) there is only KE at the end (rest) the KE is zero and particle A has gained GPE and done Work against friction by moving up the slope the unknown distnace. Have a go at writing down the tems and upload your attempt if you cant get it. You should note that the GPE and Work done against friction are similar, though GPE acts vertically and Work done against friction is along the plane. Similarly, you could use the suvat v^2 = u^2 + 2as as its equivalent to conservation of energy for simple scenarios.
Reply 2
Thank you very much for the help 
. I'm afraid I'm still having some trouble, I might have misunderstood your directions. My work goes:
Let x be the distance A moves along the surface.
The distance it moves vertically is (3/5)x, using the value of sin, so when it comes to rest it has potential energy (3/5)*2mgx.
Given it starts at rest the starting energy is 0, if we take it's height to be 0.
If we use the principle of conservation the difference between the two must equal the work done.
So (6/5)*mgx is the work done in joules, presumably against both friction and gravity?
The work done against friction and gravity is then (9/5)mgx. But then of course if we set them equal X is 0.
Obviously I'm missing something major, very sorry for all the bother. I don't know what it is about this question but I'm being stupid somewhere.
. I'm afraid I'm still having some trouble, I might have misunderstood your directions. My work goes:Let x be the distance A moves along the surface.
The distance it moves vertically is (3/5)x, using the value of sin, so when it comes to rest it has potential energy (3/5)*2mgx.
Given it starts at rest the starting energy is 0, if we take it's height to be 0.
If we use the principle of conservation the difference between the two must equal the work done.
So (6/5)*mgx is the work done in joules, presumably against both friction and gravity?
The work done against friction and gravity is then (9/5)mgx. But then of course if we set them equal X is 0.
Obviously I'm missing something major, very sorry for all the bother. I don't know what it is about this question but I'm being stupid somewhere.
Reply 3
21
Original post
by mossyrevelationsThank you very much for the help . I'm afraid I'm still having some trouble, I might have misunderstood your directions. My work goes:
Let x be the distance A moves along the surface.
The distance it moves vertically is (3/5)x, using the value of sin, so when it comes to rest it has potential energy (3/5)*2mgx.
Given it starts at rest the starting energy is 0, if we take it's height to be 0.
If we use the principle of conservation the difference between the two must equal the work done.
So (6/5)*mgx is the work done in joules, presumably against both friction and gravity?
The work done against friction and gravity is then (9/5)mgx. But then of course if we set them equal X is 0.
Obviously I'm missing something major, very sorry for all the bother. I don't know what it is about this question but I'm being stupid somewhere.
. I'm afraid I'm still having some trouble, I might have misunderstood your directions. My work goes:Let x be the distance A moves along the surface.
The distance it moves vertically is (3/5)x, using the value of sin, so when it comes to rest it has potential energy (3/5)*2mgx.
Given it starts at rest the starting energy is 0, if we take it's height to be 0.
If we use the principle of conservation the difference between the two must equal the work done.
So (6/5)*mgx is the work done in joules, presumably against both friction and gravity?
The work done against friction and gravity is then (9/5)mgx. But then of course if we set them equal X is 0.
Obviously I'm missing something major, very sorry for all the bother. I don't know what it is about this question but I'm being stupid somewhere.
There are two "suvat" phases so before and after the string breaks. So before the string breaks the acceleration is 4.48 up the slope and after the string breaks the acceleration is resolved gravity+friction down the slope which is easy enough to work out.
If you want to use conservation of energy, as the question asks, then after the string breaks
change in KE = GPE gained + work done against friction
as per the previous post for A only. The change in KE is 1/2 mv^2 where v is the velocity of A when the string breaks. Its equivalent to 0 = u^2 + 2as where the acceleration of A is given by resolved gravity+friction down the slope. The KE of A can be calculated from the motion of A and B before the string breaks though if they ask for the acceleration in the first part you could just use
v^2 = 0 + 2(4.48)1
for A and then use previous KE/GPE/Work equation.
You can do conservation of energy before string breaks, but it seems unnecessary the way the question is written.
(edited 11 months ago)
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