Question

A care is going at a speed of 21.6 km/hr when it encounters at 12.8 m long slope of angle 30° (figure 6−E5). The friction coefficient between the road and the tyre is $1/2\sqrt{3}$. 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 = 10 m/s².

Figure

Answer 1

When the driver applies hard brakes, it signifies that maximum force of friction is developed between the tyres of the car and the road.
So, maximum frictional force = μR

From the free body diagram,

R − mg cos θ = 0
⇒ R = mg cos θ (1)
and
μR + ma − mg sin θ = 0 (2)
⇒ μ mg cos θ + ma − mg sin θ = 0


where θ = 30˚
$$\begin{gathered} \Rightarrow \mathrm{\mu }g\cos \mathrm{\theta }+a-10\times \left(\frac{1}{2}\right)=0 \\ \Rightarrow a=5-\left[\frac{1}{2}\sqrt{3}\right]\times 10\left(\frac{\sqrt{3}}{2}\right) \\ =5-\frac{10\times 3}{4} \\ =\frac{20}{4}-\frac{10\times 3}{4} \\ =\frac{-10}{4} \\ =-2.5\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
s = 12.8 m
u = 6 m/s
∴ Velocity at the end of incline
$$\begin{gathered} \nu =\sqrt{u^2+2as} \\ =\sqrt{6^2+2\left(2.5\right)\left(12.8\right)} \\ =\sqrt{36+64} \\ =10\mathrm{m}/\mathrm{s}=36\mathrm{km}/\mathrm{h} \end{gathered}$$
Therefore, 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/h.