Question

A 133.5 kg crate, sliding on the floor, is brought to a stop by a 54.5 N force. What is the deceleration of the crate?

Answer 1

Answer:

0.40m/s^2

Step-by-step explanation:

Given data

Mass= 133.5kg

Force F= 54.5N

The expression for the force acting is

F=ma

substitute

54.5=133.5*a

]Divide both sides by 133.5

a= 54.5/133.5

a= 0.40 m/s^2

Hence the acceleration is 0.40m/s^2