Question

a body of mass 1000 kg is moving with the velocity of 20/s on application of breaks its come to rest 10sec find the force exerted by breaks​

Answer 1

Explanation:

Given : m=1000 kg u=10 m/s t=10 s

Applied force F=1000 N (forwards)

Retarding force f=500 N (backwards)

Acceleration of the car in forward direction a=

m

F−f

=

1000

1000−500

=0.5 m/s

2

Using v=u+at

∴ v=10+0.5×10=15 m/s