#### Question

Certain force acting on a 20 kg mass changes its velocity from 5 m s^{−1} to 2 m s^{−1}. Calculate the work done by the force.

#### Solution

Kinetic energy is given by the expression,

`(E_k)_v=1/2"mv"^2`

Where,

E_{k }= Kinetic energy of the object moving with a velocity, *v*

*m*= Mass of the object

(i) Kinetic energy when the object was moving with a velocity 5 m s^{−1}

`(E_k)_5=1/2xx20xx(5)^2=250J`

Kinetic energy when the object was moving with a velocity 2 m s^{−1}

`(E_k)_2=1/2xx20xx(2)^2=40J`

Work done by force is equal to the change in kinetic energy.

Therefore, work done by force = (E_{k})_{2} - (E_{k})_{5}

= 40 − 250 = −210 J

The negative sign indicates that the force is acting in the direction opposite to the motion of the object.

Is there an error in this question or solution?

Solution Certain force acting on a 20 kg mass changes its velocity from 5 m s^−1 to 2 m s^−1. Calculate the work done by the force. Concept: Work and Energy - Kinetic Energy.