acceleration given mass and friction

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

There are 4 forces that act on the body:

W = mg - weight of the body directed vertically down,

N = normal reaction force from the surface vertically up,

F = 600 N - force from the worker to push the crate,

f_k - kinetic friction force directed against the direction of motion.

As a result - the body moves with acceleration a and we can write the Newton's second law in vector form

mg + N + F + f_k = ma

In scalar form this equation projected on x-axis that has the same direction as a moving crate looks like

F - f_k = ma

We do not have components of weight and normal force on x-axis, because those vectors are perpendicular to the surface.

We know that friction force f_k = μ_kN and can put it in the previous equation. We get:

F - μ_kN = ma

But normal reaction force N = mg, because all other forces acts in horizontal direction. So

F - μ_kmg = ma

Finally for acceleration we have

a = (F - μ_kmg) / m

Plugging in the numbers we will obtain

a = (600 N - 0.4·100 kg·9.8 m/s²) / 100 kg = 2.1 m/s²

Answer: 2.1 m/s²