Acceleration given friction forces

Question

A worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet. The coefficients of static and kinetic friction are 0.6 and 0.4, respectively. The worker pushes with a horizontal force of 600 N. Use g = 9.8 m/s². The acceleration of the crate will be?

Answer 1

Let

m = 100 kg be the mass of the box,

μ_s = 0.6 be the coefficient of static friction,

μ_k = 0.4 be the coefficient of kinetic friction,

F = 600 N be the force pushing the box forward,

g = 9.8 m/s² be gravitational acceleration.

The weight of the box is mg.

The force of static friction

F_s = mgμ_s

The force of kinetic friction

F_k = mgμ_k

The difference

F - F_s

determines whether the worker will be able to move the crate from rest at all.

Substituting actual numbers

F - F_s = F - mgμ_s = 600 - 100×9.8×0.6 = 12 N

Since the result is positive, the worker will be able to move the box.

Once the box starts moving the effective force acting on it will be

F - F_k = F - mgμ_k

By Newton's second law, the acceleration of the crate will be

a = (F - F_k)/m = (600 - 100×9.8×0.4)/100 = 2.08 m/s²