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 0.40. 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 0.40. 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

The block will move with respect to the slab, so the coefficient of kinetic friction should be used
friction force
= 0.4* 10 * 9.81
= 39.24 N
net force acting on block
= 100 - 39.24
= 60.76 N

a) the resulting initial acceleration of the block
= 60.76 / 10
= 6.076 m/s2 ---answer

b) the resulting initial acceleration of the slab
= 39.24/ 40
= 0.981 m/s2 ---answer

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