Question

A 2.0 kg block rests on a level surface. The coefficient of static friction is µs = 0.50, and the coefficient of kinetic friction is µk = 0.30. A horizontal force, F, is applied to the block. As F is increased, the block begins moving.

⦁ Find the minimum force, F, required for the block to just start to move.
⦁ Find the force, F, required for the block to continue to move at a constant velocity.
⦁ Explain what happens to the motion of the block if a force is applied greater than those found above.

Answer 1

The minimum force, F, required for the block to just start to move is equal to the force of static friction, f s:

f s = (µ)s x N

where N is the normal force (the force perpendicular to the surface) = m x g = 2.0 kg x 9.8 m/s^2 = 19.6 N

f s = (µ)s x N = 0.50 x 19.6 N = 9.8 N

The force, F, required for the block to continue to move at a constant velocity is equal to the force of kinetic friction, f k:

f k = (µ)k x N = 0.30 x 19.6 N = 5.88 N

If a force is applied greater than the force required for the block to continue to move at a constant velocity, the block will accelerate. If the applied force is greater than the minimum force required for the block to just start to move, the block will begin moving.