How to Calculate Distance Traveled with Conservation of Energy

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A particle of mass m on an inclined slope is connected to another mass 2m hanging over a pulley, with a friction coefficient of 0.25. Initially, the particle A accelerates up the slope at 3.92 m/s², and when the string breaks after B descends 1 meter, the kinetic energy of A is calculated to be 3.92m joules. The potential energy gained by A as it moves up the slope is 5.88m joules, leading to a total energy of 9.8m joules. However, the work done against friction must be included in the energy calculations, which affects the final distance A travels after the string snaps. The corrected distance traveled by A is determined to be 0.5 meters, not the initially calculated 1.25 meters.
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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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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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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?
 
Well, numbers are pretty good except in calculation of PE terms. Since you referenced the bottom of the slope as the "zero" PE, all heights must be referenced to that point. You made in error by choosing the difference in heights betweeen initial and final PE instead. Which would be OK if you assumed the initial PE as zero, which you did not do. Solve again for x = 0.5 meters.
 
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