Question

A rightward force of 302 n is applied to a 28.6 kg crate to accelerate it across the floor. thye coefficient of friction between the crate and the floor is 0.750. determine the acceleration

Answer 1

We must all look for the fg, fn, Ff, and Fnet to get the acceleration.

 

1 Fg = 28.6 (-9.81)

= 0280.56 n

 

2 Fn = 280.56 N

 

3 Ff = μFn

= 0.750 * 280.56

Ff = 210.425 N

 

4. Fnet = Fa + Ff

= 302 – 210.42

Fnet = 91.576 N

 

5. F = ma

91.576 = 28.6a

a = 3.2 m/s