Question

A 12kg ball is given an initial velocity of 20m/s and then is rolled along the floor. A friction force of 5N opposes its motion. For what length of time will it roll before stopping. Thanks in advance!

Answer 1

a=F/m
=5/12
=0.42 m/s^2
Since the ball stops, it decelerates at a rate of 0.42 m/s^2 before stopping and it’s final velocity is 0 m/s.
t=(u-v)/a
=(20-0)/0.42
=48 s