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M2 Energy problem

This is a problem that I came across in M2. I know that I'm wrong because I checked it using the normal SUVAT method, but the question specifically asks me to use the principle of conservation of energy and I can't get the right answer using that.

"The diagram shows a particle A of mass 2m which can move on a rough surface of a plane inclined at angle x to the horizontal, where sin x = 0.6. A second particle B of mass 5m hangs freely to a light inextensible string which passes over a smooth light 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 initially hanging 2m above the ground ant A is 4m from D. When the system is released from rest with the string taut A moves up the greatest slope of the plane. When B has descended 1m the string breaks. By using the principle of conservation of energy calculate the total distance moved by A before it comes to rest."

You can probably visualise the problem from the description, D is the highest point of the triangle. This is my working.

The total potential energy lost by B by falling the 1m must be equal to the total potential energy gained by A plus the work done by A against the friction once it has reached its highest point.

The energy lost by B is mgh =5m*9.8*1=49mJ.

The work done by A against gravity is 2mgsinx*s where s is the distance travelled up the plane. The work done by A against friction is (3/8)2mgcosx*s. Simplifying (as cosx=0.8), 17.64ms=49m, s = 2.78.

However, the answer is 1.51 and I have got this result from doing this question the M1 dynamics way. What is wrong with my method? I'm assuming I've missed out an energy loss but I don't know where.
Anyone?
Reply 2
Original post by Chlorophile
Anyone?


there must be something on TV tonight

There is no one around...

I will look at your question in a while because I am doing a million things at the moment.

A brief look tells me that possibly even with energies you have to split the question into 2 different energy scenarios one before and one after the string breaks.
Which exam board?


Posted from TSR Mobile
Original post by Mutleybm1996
Which exam board?


Posted from TSR Mobile


Edexcel.
I may be wrong as I've just read this quickly, but it doesn't look like you've worked out the distance travelled correctly. I think you should work out the potential energy and kinetic energy of A at the point where the string breaks, then because KE + PE = Constant, and when A comes to rest KE = 0, you should be able to work out the potential energy when A is at (temporary) rest. From this you have a vertical height travelled and using trig you can work out the distance along the plane.
(edited 9 years ago)
Original post by Gaiaphage
I may be wrong as I've just read this quickly, but it doesn't look like you've worked out the distance travelled correctly. I think you should work out the potential energy and kinetic energy of A at the point where the string breaks, then because KE + PE = Constant, and when A comes to rest KE = 0, you should be able to work out the potential energy when A is at (temporary) rest. From this you have a vertical height travelled and using trig you can work out the distance along the plane.


Sort of, just worked out what the problem was. I forgot that the KE of particle B is also an energy loss. Factoring that in, I get the right answer. Thanks though!
Reply 7
Original post by Chlorophile
This is a problem that I came across in M2. I know that I'm wrong because I checked it using the normal SUVAT method, but the question specifically asks me to use the principle of conservation of energy and I can't get the right answer using that.

"The diagram shows a particle A of mass 2m which can move on a rough surface of a plane inclined at angle x to the horizontal, where sin x = 0.6. A second particle B of mass 5m hangs freely to a light inextensible string which passes over a smooth light 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 initially hanging 2m above the ground ant A is 4m from D. When the system is released from rest with the string taut A moves up the greatest slope of the plane. When B has descended 1m the string breaks. By using the principle of conservation of energy calculate the total distance moved by A before it comes to rest."

You can probably visualise the problem from the description, D is the highest point of the triangle. This is my working.

The total potential energy lost by B by falling the 1m must be equal to the total potential energy gained by A plus the work done by A against the friction once it has reached its highest point.

The energy lost by B is mgh =5m*9.8*1=49mJ.

The work done by A against gravity is 2mgsinx*s where s is the distance travelled up the plane. The work done by A against friction is (3/8)2mgcosx*s. Simplifying (as cosx=0.8), 17.64ms=49m, s = 2.78.

However, the answer is 1.51 and I have got this result from doing this question the M1 dynamics way. What is wrong with my method? I'm assuming I've missed out an energy loss but I don't know where.



My plan is to use energies to find the common speed of the system of 2 particles at the instant the string breaks

I got √(224/15) =2.9933 at that instant

then I am going to use energies to find how far the second particle moves up the plane
Original post by TeeEm
My plan is to use energies to find the common speed of the system of 2 particles at the instant the string breaks

I got √(224/15) =2.9933 at that instant

then I am going to use energies to find how far the second particle moves up the plane


That was the problem. For some reason, because it said conservation of energy, I stupidly thought I wasn't allowed to use SUVAT to work out velocities so didn't factor in the lost KE of B. Thanks!
Reply 9
Original post by Chlorophile
That was the problem. For some reason, because it said conservation of energy, I stupidly thought I wasn't allowed to use SUVAT to work out velocities so didn't factor in the lost KE of B. Thanks!


all good:smile:

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