Answer to Question #298322 in Physics for Kkmac

Question

A crate at rest with a mass of 75 kg is being pushed by a force of 620 N. If the coefficient of kinetic friction between the crate and the floor is 0.55, determine how fast the crate will accelerate, and how far the crate will travel in 3.0 seconds.

Answer 1

Given:

$m=45\:\rm kg$

$F=620\:\rm N$

$\mu_k=0.55$

$t=3.0\:\rm s$

The Newton's second law says

$$a=\frac{F_{net}}{m}=\frac{F-\mu_k mg}{m}$$

$$a=\frac{620-45*9.8*0.55}{45}=8.4\:\rm m/s^2$$

The distance traveled by a crate

$$d=at^2/2=8.4*3.9^2/2=38\:\rm m$$