#### joker_900

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OK there is a slope inclined at an angle z to the horizontal, where z=arcsin0.6. There is a particle of mass m on the slope (A) attached to a string which goes over a pulley at the top corner and to another particle of mass 2m (B)hanging over the side. Coefficient of friction = 0.25.

In the first part i worked out that A accelerates up the slope at 3.92ms^-2

B descends 1m and then the string breaks. Use conservation of energy to find the total distance A moves before coming to rest.

So i thought, use v^2=u^2 + 2as to show that v^2 = 7.84 when the string breaks, and so the KE of A when the string breaks is 3.92m joules. Then as it goes 1m up the slope, it gains 5.88m joules of PE (using trigonometry). So when the string breaks A has 9.8m joules of energy.

Work done against friction = 0.25*mgcosz * x = 1.96mx joules (where x is the distance A travels up the slope after the string snaps)

So surely:

9.8m - 1.96mx = mgh (as at rest, A has only PE)
9.8m - 1.96mx = 0.6xmg
x = 1.25 metres

So total distance = 2.25 metres

But the answer is 1.5 metres!

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#### berkeman

Mentor
You need to include the heat energy lost due to friction. Energy is force multiplied by distance. Do you see the extra term you need to put in?

EDIT -- I see that you are trying to include the frictional term (sorry I missed that in my first read). Give me a minute here...

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#### berkeman

Mentor
I have to run to a meeting, but I'm getting a different net acceleration up the slope before the string breaks. I'm getting a number a little over g. (I could be wrong of course). Can you explain the 3 components of the net acceleration during that time?