Question

Question 4
Certain force acting on $20kg$ mass changes its velocity from $5m{s}^{-1}$ to $2m{s}^{-1}$. Calculate the work done by the force.

Answer 1

Kinetic energy is given by the expression,$(E_k{)}_v=\frac{1}{2}m{v}^2$
Where,
$E_K$= Kinetic energy of the object moving with a velocity, v

(i) Kinetic energy when the object was moving with a velocity $5m{s}^{-1}$
$(E_k{)}_5=\frac{1}{2}\times 20\times (5{)}^2=250J$

(ii) Kinetic energy when the object was moving with a velocity $2m{s}^{-1}$
$(E_k{)}_2=\frac{1}{2}\times 20\times (2{)}^2=40J$

The work done by the force = $(E_k{)}_2-(E_k{)}_5$
= $40J-250J$
= $-210J$
The negative sign indicates the retarding nature of the applied force.