# Certain force acting on a 20 kg mass changes its velocity from 5 m s power of minus 1 to 2 m s power of minus 1 Calculate the work done by the force

## Mass.m = 20 kg.

Initial velocity, u=5m/s

Final velocity, v=2m/s

Work done = change in K.E.

$W = \frac{1}{2}m u^2 -\frac{1}{2}m v^2 = \frac{1}{2}m( u^2- v^2)$

$= \frac{1}{2}20( 2^2- 5^2) = -210J$

the negative sign implies the retarding

nature of the applied force.

## Related Chapters

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-