The figure below shows a box of mass m=2.0 kg on a ramp of angle ?=30 degrees ti

Question

The figure below shows a box of mass m=2.0 kg on a ramp of angle ?=30 degrees tied to a light rope passing over a light pulley and attached to an ideal spring with constant k=70 N/m. The rope does not stretch or break. The box is released from rest when the spring is unstretched. Once released, the box travels downward a distance x=20 cm, where it stops and stays at rest. The coefficient of static friction between the ramp and the block is µs and the coefficient of kinetic friction is µk . These coefficients are not given.

a) Use the work-energy theorem and the information provided to calculate the coefficient of kinetic friction µk.

b) What is the magnitude and direction of the force of friction at the farthest point x=20 cm?

Explanation / Answer

a) on block perpendicular to the incline,

N - mgcos@ = 0

N = mgcos@

and friction force = uk N = uk mg cos@

using work energy theorem,

work done by gravity + work done by spring + work done by friction = change in KE

mgxsin@ - kx^2/2 - ukxmgcos@ = 0 - 0

2 x 9.81 x 0.20 x sin30 - (70 x 0.20^2 /2) - (uk x 0.20 x 2 x 9.81 x cos30) = 0

3.40 uk = 0.562

uk = 0.165 ...........Ans

b) at this point x = 0.20 m balancing forces along incline,

kx - mgsin30 - ukmgcos30 = 0

70 x 0.20 = 2 x 9.81 x sin30 + uk x 2 x 9.81 x cos30

uk = 0.247 ............Ans

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