A particle can slide along a track with elevated ends and a flat central part, a

Question

A particle can slide along a track with elevated ends and a flat central part, as shown below. The flat part has length L = 40 cm. The curved portions of the track are frictionless, but for the flat part the coefficient of kinetic friction is mu_k = 0.30. The particle is released from rest at point A, which is a height h = L/2. How far from the left edge of the flat does the particle finally stop? cm In the figure below, the pulley has negligible mass, and both it and the inclined plane are frictionless. Block A has a mass of 1.0 kg, and block B has a mass of 2.0 kg, and angle theta is 30degree. If the blocks are released from rest with the connecting cord taut, what is their total kinetic energy when block B has fallen 22 cm? J

Explanation / Answer

1) Potential energy = work against friction .
mgL/2 = F* s
mgL/2 =mg* s
L/2 = s

s = 40/(2 *0.3) = 66.66

But the horizontal path is only 40 cm
There fore it returns from the right side and stops at 26.66 cm from the right .
Or 13.34 cm from the left.

2)

Potential energy loss of block B = mB*g*h

If block B moves h down, block A moves h up along the plane. The change in elevation of block A is then h*sin and it gains potential energy of mA*g*h*sin.

The net loss of potential energy is then

PE(loss) = mB*g*h - mA*g*h*sin

for h = 0.22 m
mA = 1kg
mB = 2 kg
= 30°

KE = PE loss = 3.23 J

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