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# Stuck on M2 work question watch

1. A particle of mass 2m can move on the rough surface of a plane inclined at an angle theta to the horizontal where sintheta=3/5 . A second particle of mass 3m hangs freely attached to a light inextensible string which passes over a smooth pulley fixed at D. The other end of the string is attached to the 2m mass on the inclined plane. The coefficien of friction is 0.25. The system is released from rest with the string taut and the 2m mass moves up a line of greatest slope of the plane. When each particle has moved a distance s, the 2m mass has not reached the pulley and the 3m mass has not reached the ground.

A) find an expression for the potential lost by the system when each particle has moved a distance of s.
Energy lost=9mgs/5
B) When each particle has moved a distance of s, they are moving at a speed v. Find an expression for v2 in terms of s.
I have found so far:
a=2.94 ms^-2
Friction= 3mg/10

What do I do now?
2. (Original post by bobbricks)
A particle of mass 2m can move on the rough surface of a plane inclined at an angle theta to the horizontal where sintheta=3/5 . A second particle of mass 3m hangs freely attached to a light inextensible string which passes over a smooth pulley fixed at D. The other end of the string is attached to the 2m mass on the inclined plane. The coefficien of friction is 0.25. The system is released from rest with the string taut and the 2m mass moves up a line of greatest slope of the plane. When each particle has moved a distance s, the 2m mass has not reached the pulley and the 3m mass has not reached the ground.

A) find an expression for the potential lost by the system when each particle has moved a distance of s.
Energy lost=9mgs/5
B) When each particle has moved a distance of s, they are moving at a speed v. Find an expression for v2 in terms of s.
I have found so far:
a=2.94 ms^-2
Friction= 3mg/10

What do I do now?
For (B) Use gain of kinetic energy=loss of potential energy - work done againbst friction.
i.e. the work-energy principle
3. (Original post by brianeverit)
For (B) Use gain of kinetic energy=loss of potential energy - work done againbst friction.
i.e. the work-energy principle
I did 1/2 (2m)v^2=9(2m)gs/5 -2(2m)gs/5
but that doesn't give the correct answer??

EDIT: I got v^2 =14gs/5 which I think is correct..?
4. (Original post by bobbricks)
I did 1/2 (2m)v^2=9(2m)gs/5 -2(2m)gs/5
but that doesn't give the correct answer??

EDIT: I got v^2 =14gs/5 which I think is correct..?
what about the gain in k.e. of the mass 3m?

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Updated: January 8, 2015
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