Question

A person pushes horizontally with a force of 220. N on a 56.0 kg crate to move it across a level floor. The coefficient of kinetic friction is 0.21. (a) What is the magnitude of the frictional force? (b) What is the magnitude of the crate's acceleration?

Answer 1

Step-by-step explanation:

(a) As it is given that there is no vertical displacement. Hence, the forces on y-axis are equal and opposite.

Also, it is known that F = mg

So, F =
`56 kg * 9.8 m/s^(2)`

= 548.8 N

Also,
`F_(f) = \mu N`

where,
`\mu` = coefficient of kinetic friction

N = force calculated

Putting the values into the formula as follows.

`F_(f) = \mu N`

=
`0.21 * 548.8`

= 115.25 N

Hence, the magnitude of the frictional force is 115.25 N.

(b) Now, let us assume that the acceleration is towards the positive x-direction.

`F - F_(f) = ma`

`548.8 N - 115.25 = 56 * a`

a = 7.74
`m/s^(2)`

Therefore, the magnitude of the crate's acceleration is 7.74
`m/s^(2)`.