Question

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is the acceleration of the crate if the 750-N force is maintained after the crate begins to move and the coefficient of kinetic friction is 0.12?

Answer 1

Answer:

The acceleration of the crate is $1.82\ m/s^2$.

Explanation:

Given that,

Force, F = 750 N

Mass of the crate, m = 250 kg

The coefficient of friction is 0.12.

We need to find the acceleration of the crate. The net force acting on the crate is given by :

$F=ma\\\\F-f=ma$

f is frictional force, $f=\mu N=\mu mg$

$F-\mu mg=ma\\\\a=\dfrac{F-\mu mg}{m}\\\\a=\dfrac{750-0.12\times 250\times 9.8}{250}\\\\a=1.82\ m/s^2$

So, the acceleration of the crate is $1.82\ m/s^2$. Hence, this is the required solution.