Question

Find the average frictional force needed to stop a car weighing 500 kg at a distance of 25 m if the initial speed is 72 km/h.

Answer 1

The frictional force is 4000 N.

              Explanation:

Here Given :-

mass ( m ) = 500 kg

Initial speed u = 75 km/h

Converting kilometre per hour to meter per second

(72000 m) / (3600 s) = 20 m/s

u = 20 m/s.

Distance (s)  = 25 m

Final speed (v)  = 0

We know the formula

$\begin{equation}v^{2}-u^{2}=2 a s\end$

on putting the value

$\begin{equation}\begin{aligned}&\begin{aligned}0^{2}-20^{2} &=2 \times a \times 25 \\400 &=50 a\end{aligned}\\&a=\frac{400}{50}\\&a=8 \mathrm{m} / \mathrm{s}^{2}\end{aligned}\end$

But then the acceleration is the other way around

a = -8 m/s²

So Now the frictional force F =  m × a

F = 500 × 8

F = 4000 N.

The frictional force is therefore 4000 N.                

Answer 2

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The frictional force is 4000 N