Question

A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle ${30}^{\circ }$ (figure 6-E5).
The friction coefficient between the road and the tyre is \frac{1}{2\sqrt{3}}. Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than 36 km/hr. Take g = $10m/s^2$

Answer 1

Hardest brake indicates that maximum force of friction s developed between car's tyre and road.
Maximum frictional force $=\mu R.$
From the free body diagram,
$R-mg\cos \theta =0$
$\Rightarrow r=mg\cos \theta and\mu R+ma-mg\sin \theta =0\Rightarrow \mu mg\cos \theta +ma-mgsin\theta =0$

$mg\cos \theta +a-10\times \left(\frac{1}{2}\right)=0\Rightarrow a=5-[1-2\sqrt{3}]\times 10\left(\frac{\sqrt{3}}{2}\right)=5-2.5=2.5m/s^{2}S=12.8m,u=6m/s$
$\therefore$ Velocity at the end of incline
$v=\sqrt{u^2+2as}=\sqrt{6^2+2\left(2.5\right)\left(12.8\right)}=\sqrt{36+64}10m/s=36km/hr$
Hence, the harder the driver applies the brakes, the lower will be the velocity of the car when it reaches the ground; i.e. at 36 km/hr.