Question

A car of mass 1100 kg is moving on a straight rough horizontal surface with a speed of 43.2 kmph. If the coefficient of friction between the surface and the tyres of the car is 0.3, then the shortest distance in which the car can come to rest is (Acceleration due to gravity = $10~\text{m s}^{-2}$)

• A. 24 m
• B. 48 m
• C. 12 m
• D. 36 m

Hint:

Always convert speed to SI units before using in equations. Use work-energy principle for friction problems: $\text{Kinetic energy lost = Work done by friction}$.

Answer 1

The Correct Option is B

1. Convert speed to m/s: $v = 43.2~\text{kmph} = 43.2 \times \frac{1000}{3600} = 12~\text{m s}^{-1}$.
2. Frictional force: $f = \mu m g = 0.3 \times 1100 \times 10 = 3300~\text{N}$.
3. Using work-energy principle: $\frac{1}{2} m v^2 = f \cdot d$
$\frac{1}{2} \cdot 1100 \cdot 12^2 = 3300 \cdot d \implies 79200 = 3300 d \implies d = 24~\text{m}$