Question

A car of mass $1000kg$ is moving with a speed of $90km{h}^{-1}$ to stop it in $5seconds$, calculate the force applied by brakes and total work done.

Answer 1

Step 1: Given data

$$\begin{gathered} \mathrm{mass}\mathrm{of}\mathrm{car},\mathrm{m}=1000\mathrm{kg} \\ \mathrm{Initial}\mathrm{velocity},\mathrm{u}=90\mathrm{km}/\mathrm{hr}=90\times \frac{5}{18}\mathrm{m}/\sec =25\mathrm{m}/\sec \\ \mathrm{Stop}\mathrm{time},\mathrm{t}=5\mathrm{s} \end{gathered}$$

Step 2: Find the acceleration

$\mathrm{Final}\mathrm{velocity},\mathrm{v}=0$

Using the first equation of motion

$$\begin{gathered} v=u+at \\ 0=25+a\times 5 \\ a=-5m/se{c}^2\left(retardation\right) \end{gathered}$$

Step 3: Find the force

The force applied by a break

$$\begin{gathered} Force=ma \\ Force=1000\times (-5m/se{c}^2) \\ Force=-5000N \end{gathered}$$

Negative force implies the force acting on an object against the applied force.

Step 4: Find the distance

Distance traveled to stop the car after applying the break is s.

Using the third equation of motion

$$\begin{gathered} v^2=u^2+2as \\ 0={25}^2-2\times 5\times s \\ 0=625-10s \\ s=\frac{625}{10} \\ s=62.5m \end{gathered}$$

Step 5: Find the work done

$$\begin{gathered} W=Fs \\ W=-5000\times 62.5 \\ W=-312.5KJ \end{gathered}$$

Here the negative sign indicates that the work is done against the displacement.

Hence, the work done is $\mathbf{312}\mathbf{.}\mathbf{5}\mathit{K}\mathit{J}$.