Suppose a body of mass 10kg is initially moving with velocity 15m/s on ground. Because of friction from ground it stops after some time. Find the work done by frictional force.

Open in App

Solution

Workdone = KE(final)-KE (initial)
= 1/2 * 10 * 0^2 -(1/21015^2)
= -1125 j

Suggest Corrections

Similar questions

Q. A body of mass

10 k g

is moving on a horizontal surface by applying a force of

10 N

in forward direction. If body moves with constant velocity, the work done by force of friction for a displacement of

2 m

is :

Physics says that a body after travelling a short distance stops because of frictional force but according to some a body starts moving because of applied force to it and stops because the applied energy exhausts after travelling a certain distance.so which is true and explain why.

Q. A body is rolling on the ground with a velocity of

1 m / s

. After travelling a distance of

5 m

, the body stops. The coefficient of friction is:

Q. A block of mass 2 kg slides on a rough surface at t=0, if speed is 2 m/s. It stops after covering a distance of 20 cm because of friction. Find work done by the friction.

A body of mass 10 kg is moving on a circular track of radius 15 m. What is work done by the centripetal force on it in one complete revolution? (Take g = $10 m / s^{2}$ )

Join BYJU'S Learning Program

Suppose a body of mass 10kg is initially moving with velocity 15m/s on ground. Because of friction from ground it stops after some time. Find the work done by frictional force.

Workdone = KE(final)-KE (initial) = 1/2 * 10 * 0^2 -(1/2*10*15^2) = -1125 j

Workdone = KE(final)-KE (initial)
= 1/2 * 10 * 0^2 -(1/21015^2)
= -1125 j