Question

A car having a mass of 1 metric ton is moving with a speed of 30 m/s. It suddenly applies the brakes and skids to rest in a certain distance d. The frictional force between the tyres and road is 6000 N. What is the value of d?

• A. 50 m
• B. 25m
• C. 35m
• D. 75m

Answer 1

Answer: (D)

75 m

Answer 2

The car's kinetic energy is dissipated by the work done by the frictional force.

1. Identify Given Values:

   • Mass of the car, $m = 1 \text{ metric ton} = 1000 \text{ kg}$
   • Initial speed, $u = 30 \text{ m/s}$
   • Final speed, $v = 0 \text{ m/s}$ (since it comes to rest)
   • Frictional force, $F_f = 6000 \text{ N}$

2. Apply the Work-Energy Theorem:

   The work done by the frictional force is equal to the change in the car's kinetic energy.

   Work done by friction ($W$) = $F_f \times d$, where $d$ is the distance skidded.

   Change in kinetic energy ($\Delta KE$) = $KE_{final} - KE_{initial}$

   $KE_{initial} = \frac{1}{2}mu^2$

   $KE_{final} = \frac{1}{2}mv^2 = \frac{1}{2}m(0)^2 = 0$

   According to the Work-Energy Theorem:

   $W = \Delta KE$

   The work done by friction is negative because the force opposes displacement, but we are considering the magnitude of work done to stop the car. So, the magnitude of work done by friction equals the initial kinetic energy.

   $F_f \times d = \frac{1}{2}mu^2$

3. Calculate the Distance (d):

   Rearrange the formula to solve for $d$:

   $d = \frac{\frac{1}{2}mu^2}{F_f}$

   Substitute the given values:

   $d = \frac{\frac{1}{2} \times 1000 \text{ kg} \times (30 \text{ m/s})^2}{6000 \text{ N}}$

   $d = \frac{500 \text{ kg} \times 900 \text{ (m/s)}^2}{6000 \text{ N}}$

   $d = \frac{450000 \text{ J}}{6000 \text{ N}}$

   $d = 75 \text{ m}$

The value of d is 75 m.