Question

A force is applied on a body of mass $20kg$ moving with a velocity of $40m{s}^{-1}$.The body attains a velocity of $50m{s}^{-1}$in $2s$.Calculate the work done by the body.

Answer 1

Step 1: Given data:

Mass, $m=20kg$

Initial velocity, $u=40m{s}^{-1}$

Final velocity, $v=50m{s}^{-1}$

Time, $t=2s$

Step 2: Derivation of formula to find the work done by initial velocity, final velocity, mass, and time:

The first equation of motion, $v=u+at$ $\left(1\right)$

Where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration and $t$ is time.

From $\left(1\right)$, acceleration is rewritten as,

$a=\frac{v-u}{t}$ $\left(2\right)$

Similarly, we know the formula for force,

$F=m.a$ $\left(3\right)$

Where $F$ is force, $a$ is acceleration and $m$ is mass

Substitute equation $\left(2\right)$ in equation $\left(3\right)$ we get,

$F=m\left(\frac{v-u}{t}\right)$ $\left(4\right)$

Using the third equation of motion, displacement is given by the formula,

Displacement $=\frac{(v^2-u^2)}{2a}$ $\left(5\right)$

We know that,

Work $=$ Force $\times$ displacements

Substitute equation $\left(3\right)$ and $\left(5\right)$ in above equation we get,

$W=ma\times \frac{(v^2-u^2)}{2a}$

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

Step 3: Calculation of work done:

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

$W=\frac{1}{2}\times 20\times ({50}^2-{40}^2)$

$=10\times (2500-1600)$

$=10\times 900$

$W=9000J$

Therefore, the work done to attain the final velocity is $9000J$.