Question

The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of motion. Prove that the increase in kinetic energy of the body is equal to the work done by the force on the body.

Answer 1

Step 1: Work done by force on body is given by the equation as follows:

$\mathrm{W}=\mathrm{f}\times \mathrm{s}$ …… (1)

Here, $\mathrm{W}$ is work done, $\mathrm{f}$ is force and $\mathrm{s}$ is displacement of body.

Step 2: Velocity of a body moving is given by the equation as follows:

${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as}$

Here, $\mathrm{v}$ is final velocity, $\mathrm{u}$ is initial velocity, $\mathrm{a}$ is acceleration due to gravity and $\mathrm{s}$ is displacement of body.

Rearrange the equation for displacement as follows:

$\mathrm{s}=\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2\mathrm{a}}$ …… (2)

Step 3: Acceleration produced on a body on applying force is given by the equation as follows:

$\mathrm{a}=\frac{\mathrm{f}}{\mathrm{m}}$

Here, $\mathrm{a}$ is acceleration, $\mathrm{f}$ is force and $\mathrm{m}$ is mass of body.

Rearrange the equation for force as follows:

$\mathrm{f}=\mathrm{m}\times \mathrm{a}$ …… (3)

Step 4: Substitute equations (2) and (3) in equation (1) as follows:

$\mathrm{W}=\mathrm{m}\times \mathrm{a}\times \left(\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2\mathrm{a}}\right)$

$\mathrm{W}=\mathrm{m}\times \left(\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2}\right)$

$\mathrm{W}=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mu}}^2$

Here, ${\mathrm{KE}}_{\mathrm{final}}=\frac{1}{2}{\mathrm{mv}}^2$ and ${\mathrm{KE}}_{\mathrm{initial}}=\frac{1}{2}{\mathrm{mu}}^2$

Thus,

$\mathrm{W}={\mathrm{KE}}_{\mathrm{final}}-{\mathrm{KE}}_{\mathrm{initial}}$

Thus, work done is the increase in kinetic energy i.e. the difference in kinetic energy.