A block of mass 2.20 kg is accelerated across a rough surface by a light cord pa

Question

A block of mass 2.20 kg is accelerated across a rough surface by a light cord passing over a small pulley as shown in the figure below. The tension T in the cord is maintained at 10.0 N, and the pulley is 0.120 m above the top of the block. The coefficient of kinetic friction is 0.360.

a) Determine the acceleration of the block when x = .400 m.

b) Find the maximum value of the acceleration and the position x for which it occurs.


x = m
a = m/s2


(c) Find the largest value of x for which the acceleration is zero.

Explanation / Answer

So the block is being pulled along by a force of 10N minus kinetic friction. The force of kinetic friction is the normal force multiplied by the coefficient of kinetic friction. Note that in this case the normal force is the weight of the object MINUS the lift you get from the pulley being above the block. The amount of lift depends on the location of the block because as you get closer the angle of the cord gets higher and there is less of the mass "pushing down" on the rough surface. At the same time, the lateral component of the tension in the cord is decreasing. See if that's enough to get you started setting up the free-body diagram and the forces. It's really not that hard once you start from there. Intuitively, the maximum value of the acceleration occurs where the lateral pull is highest and the normal force is lowest. That should be somewhere near 45° on the cord, although not exactly.

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