Certain force acting on a 20 kg mass changes its velocity from `5ms^(-1)` to `2ms^(-1)`. Calculate the work done by the force.

Answer

The work done by the force is `-210J`
(The negative sign indicates the retarding nature of the force applied).

Explanation

Here, `m=20kg, u=5ms^(-1), v=2ms^(-1),W=?`
The work done by the force
= change in kinetic energy
= Final `E_(k)` - Initial `E_(k)`
`=(1)/(2)mv^(2)-(1)/(2)m u^(2)`
`=(1)/(2)m(v^(2)-u^(2))`
`=(1)/(2)m(v^(2)-u^(2))`
`=(1)/(2)xx20xx[(2)^(2)-(5)^(2)]`
`=10xx(4-25)=10xx(-21)=-210J`