Question

A person pushes horizontally with a force of 170. N on a 66.0 kg crate to move it across a level floor. The coefficient of kinetic friction is 0.150. (a) What is the magnitude of the frictional force? (b) What is the magnitude of the crate's acceleration? Use g=9.81 m/s2.

Answer 1

(a)
`f_k=97.1N`

(b)
`a=1.10m/s^(2)`

Step-by-step explanation:

First, we write the equations of motion of the crate for each axis:

`x: F-f_k=ma\\\\y:N-mg=0`

Since the kinetic frictional force is equal to
`\mu N`, and from the second equation we have that:

`f_k=\mu N=\mu mg\\\\f_k=(0.150)(66.0kg)(9.81m/s^(2))=97.1N`

This means the frictional force has a magnitude of 97.1N (a).

Next, we use this value to calculate the magnitude of acceleration from the first equation of motion:

`a=(F-f_k)/(m) \\\\a=(170N-97.1N)/(66.0kg)=1.10m/s^(2)`

In words, the magnitude of the crate's acceleration is 1.10m/s² (b).