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 the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

Answer 1

Step 1: Given data

A constant force $F$

Step 2: To prove

The increase in the kinetic energy of the body is equal to the work done by the force on the body.

$∆K.E.=W$, where $∆K.E.$ is the change in kinetic energy and $W$ is the work done.

Step 3: Formulae used

$W=F\times s$, where $F$ is the force and $s$ is the displacement…….(1)

$F=ma$, where $m$is the mass and $a$ is the acceleration………(2)

$K.E.=\frac{1}{2}m{v}^2$, where $v$ is the velocity.

Step 4: Proof of the mentioned statement:

Using the third equation of motion, we know that $v^2-u^2=2as$, where $v$ is final velocity and $u$ is initial velocity.

Rearranging this equation of motion, we get $s=\frac{v^2-u^2}{2a}$.

Substituting the value of $s$ and $F$ in formula 1 from the above expression and formula 2 respectively,

$$\begin{gathered} W=ma\times \frac{v^2-u^2}{2a} \\ W=\frac{m}{2}\left(v^2-u^2\right) \\ W=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2 \end{gathered}$$

Again, substituting values from formula 3 into the above expression,

$$\begin{gathered} W=K.E{.}_f-K.E{.}_i \\ W=∆K.E. \end{gathered}$$

Hence. it is proved that the increase in the kinetic energy of the body is equal to the work done by the force on the body.