Question

A $1500kg$ car is moving at a speed of $50m{s}^{-1}$ When brakes are applied it stops with uniform retardation at a distance of $150m$. Calculate the force applied by the brakes of the car and the work done before stopping?

Answer 1

Step 1: Given data

$$\begin{gathered} Mass,\mathrm{m}=1500\mathrm{kg} \\ Initialvelocity,\mathrm{u}=50\mathrm{m}/\mathrm{s} \\ displacement,\mathrm{s}=150\mathrm{m} \\ Finalvelocity,\mathrm{v}=0 \end{gathered}$$

Step 2: Find the acceleration

Using the third equation of motion

$$\begin{gathered} v^2=u^2+2as \\ 0={50}^2+2\times a\times 150 \\ a=\frac{-2500}{300} \\ a=\left(\frac{-25}{3}\right)m/s^2 \end{gathered}$$

Step 3: Find the force

$$\begin{gathered} \mathrm{Force},\mathrm{F}=\mathrm{ma} \\ \mathrm{F}=1500\times \left(\frac{-25}{3}\right) \\ \mathrm{Force}=-12500\mathrm{N} \end{gathered}$$

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

Step 4: Finding the work done

$$\begin{gathered} \mathrm{Work}\mathrm{done},\mathrm{W}=\mathrm{F}.\mathrm{s} \\ \mathrm{W}=-12500\times 150 \\ \mathrm{W}=-1875\times {10}^3\mathrm{J} \\ \mathrm{W}=-1875\mathrm{KJ} \end{gathered}$$

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

Hence, the force applied is $12500N$ and work done is $\mathbf{-}\mathbf{1875}\mathbf{}\mathit{k}\mathit{J}$.