Question

A 690.0 N force is applied on a 130.0 kg desk to move it across the floor. The coefficient of friction between the crate and the floor is 0.31. What is the acceleration of the crate?

Answer 1

Based on the Newton second law, you can write the following equation:

`F-F_r=ma`

where,

F: applied force = 690.0N

Fr: friction force = μN

μ: coefficient of friction = 0.31

N: normal force = m*g

m: mass of the desk = 130.0kg

g: gravitational acceleration constant = 9.8m/s^2

Solve the equation above for a, replace the previous values of the parameters and simplify, as follow:

Hence, the acceleration of the crate is 295.06m/s^2