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°. The friction coefficient between the road and the tyre is 1/2√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.

Answer 1

Answer:

See below.

Explanation:

Observe the attached diagram.

We can solve the problem with the help of Work Energy Theorem and Principle of Conservation of Energy.

The car is moving at a velocity of 21.6 km/hr which is 6 m/s. It's kinetic energy will be,

$K=\dfrac {1}{2}M\left( 6\right) ^{2}=18M$

When the car moves down the rough slope, work is sone by the frictional force. It will be given by,

$\begin{aligned}W_{f}=\left( 12.8\right) \mu \ Mg\cos \theta \\ =\left( 12.8\right) \dfrac {1}{2\sqrt {3}}M\left( 10\right) \dfrac {\sqrt {3}}{2}\\ =32M\end{aligned}$

The height of the slope is,

$\begin{aligned}h=\left( 12.8\right) \sin 30^{\circ }\\ =6.4m\end{aligned}$

The car also loses its height while descending. Therefore, change in potential energy,

$\begin{aligned}P=Mgh\\ =M\left( 10\right) \left( 6. 4\right) \\ =64M\end{aligned}$

Now, we need to equate all the above energies. First, the car had its know kinetic energy ( K ), then by descending through a height it gained potential energy ( P ). The friction opposed the motion and hence some energy was lost ( Wf ).

$\begin{aligned}E=K+P-W_{f}\\ =18M+64M - 32M\\ =50M\end{aligned}$

This is the kinetic energy of the car when it reaches the bottom. Its velocity will be,

$\begin{aligned}50M=\dfrac {1}{2}Mv^{2}\\ v=10 \ m \ s^{-1}\\ =36 \ km \ hr^{-1}\end{aligned}$

which gives us the answer.

Tip: I used to get worried when the mass of the car wasn't provided. Because if mass is not provided, we need to solve the question with kinematical equations, in some cases. In most WE questions, the term accounting for mass cancels out in the last expression.