Mechanics problem

yonski · 2007-06-07 03:31:11

Hi there,
i had a mechanics exam this morning and i was struggling with this questions (the numbers were maybe a bit different, but it was basically the same). I'm worried that i've messed it up, so was wondering whether someone could go over my working please to put my mind at rest?

Okay, so you have 2 particles attached to a taut string which passes over a smooth pulley, as in the diagram below:

Particle A rests on the slope, and the coefficient of friction between the slope and the particle is 4/7 . Also, tan(x) = 3/4 .

(a) The system is released from rest and the two particles move a distance h (A does not reach the pulley and B does not reach the floor). Calculate the potential energy lost by the system as a whole when the particles have moved a distance h.

(b) When the particles have moved distance d, their speed is v ms/s, where v^2 = kgh . Work out the value of k.

My solutions:

(a) In moving a distance h up the slope, A gains potential energy (3/5)mgh joules.

In moving downwards a distance h, B loses potential energy 2mgh joules.

Therefore net energy loss is: 2mgh - (3/5)mgh = (7/5)mgh joules.

(b) Some of this potential energy lost goes towards work done against friction, and some gets converted into kinetic energy.

Work done against friction = (4/7)mghcos(x) = (16/35)mgh joules.

Therefore kinetic energy = (7/5)mgh - (16/35)mgh = (33/35)mgh joules.

Hence the speed of the particles is v, where

0.5*3mv^2 = (33/35)mgh (This is the bit i'm unsure about - do you take the mass as the sum of the two?)

(3/2)v^2 = (33/35)gh

v^2 = (22/35)gh .

Therefore k = 22/35.

Is that okay, or have i made some mistakes? Any help appreciated

Last edited by yonski (2007-06-07 03:34:33)

John E. Franklin · 2007-06-08 00:08:03

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#1 2007-06-07 03:31:11

Mechanics problem

#2 2007-06-08 00:08:03

Re: Mechanics problem

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