# M2 Energy problem

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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.

"

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.

"

*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.

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#3

(Original post by

Anyone?

**Chlorophile**)Anyone?

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.

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#6

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.

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(Original post by

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.

**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.

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#8

(Original post by

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.

"

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.

**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

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(Original post by

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

**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

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#10

(Original post by

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!

**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!

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