In the figure below, a slab of mass m 1 = 40 kg rests on a frictionless floor, a

Question

In the figure below, a slab of mass m1 = 40 kg rests on a frictionless floor, and a block of mass m2 = 10 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.30. The block is pulled by a horizontal force with a magnitude of 100 N.

In the figure below, a slab of mass m1 = 40 kg rests on a frictionless floor, and a block of mass m2 = 10 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.30. The block is pulled by a horizontal force with a magnitude of 100 N. What is the resulting acceleration of the block? What is the resulting acceleration of the slab?

Explanation / Answer

normal force on slab by block
= 10 * 9.81 N
friction force
= 0.5* 10 * 9.81
= 49.05 N
net force acting on block
= 100 - 49.05
= 50.95 N

a) the resulting initial acceleration of the block
= 50.95 / 10
= 5.095 m/s2 ---answer

b) the resulting initial acceleration of the slab
= 49.05 / 40
= 1.226 m/s2 ---answer

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