A 3.00 kg block slides from rest from the bottom of an inclined plane, tilted up

Question

A 3.00 kg block slides from rest from the bottom of an inclined plane, tilted up at 25 degrees. This acceleration is due to a tied string that goes up over a pulley attached to a 2.00 kg mass that falls from rest 2.00 m to the floor.

(a) How fast is the block traveling up the ramp when the falling block hits the floor if friction on the ramp is negligible?

(b) How fast is the block traveling up the ramp when the falling block hits the floor if the coefficient of friction on the ramp is 0.20? Use conservation of energy techniques.

I am REALLY struggling with this and all previous answers seem to be wrong in some way and I cant find my way around this problem... PLEASE help and explain so that I can understand the concepts? Here is a diagram I made:

Explanation / Answer

a) when 2 kg mass falls to the ground it looses potential energy = mgh = 2*9.8*2 = 39.2 3kg block moves up the ramp by a distance = 2 m so 3kg is at a height of 2sin(25) from the ground 3kg block will gain some PE = 3*9.8*2sin(25) = 24.85 Net loss in PE = 39.2-24.85 = 14.35 This loss is compensated by the gain in kE of both the masses. Since they are tied by a same string they will have same speed .5*2*v^2+.5*3*v^2 = 14.35 solving we get v = 2.4 m/s b) Here we have to take in the energy loss due to friction force also Friction force = .2*3*9.8*cos(25) = 5.33 energy lost due to friction force = 5.33*2 = 10.66 Here energy available for the increase in KE of the masses = 14.35-10.66 .5*2*v^2+.5*3*v^2 = 14.35-10.66 solving we get v = 1.215 m/s I hope this is clear ..if you need any clarification please let me know

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