Question

A car moving with a speed \( v \) is stopped at a distance \( d \) by a retarding force \( F \). The force needed to stop the same car moving with the speed \( 3v \) within the same distance is:

• A. \( 3F \)
• B. \( 6F \)
• C. \( 8F \)
• D. \( 9F \)
• E. \( 12F \)

Hint:

When stopping a car, the required stopping force is proportional to the square of the velocity. Therefore, if the velocity increases by a factor of 3, the required force increases by a factor of \( 3^2 = 9 \).

Answer 1

The Correct Option is D

The work-energy theorem states that the work done by the retarding force is equal to the change in the kinetic energy of the car. \[ W = \Delta KE = \frac{1}{2} m v^2 \] where \( m \) is the mass of the car and \( v \) is the initial velocity. The work done by the force \( F \) over a distance \( d \) is: \[ W = F \times d \] Thus, we have the equation: \[ F \times d = \frac{1}{2} m v^2 \] For the car moving with a speed \( 3v \), its initial kinetic energy will be: \[ KE_{{new}} = \frac{1}{2} m (3v)^2 = \frac{1}{2} m \times 9v^2 = 9 \times \left( \frac{1}{2} m v^2 \right) \] Since the car is to stop within the same distance \( d \), the work done by the retarding force should equal this new kinetic energy: \[ F_{{new}} \times d = 9 \times \left( F \times d \right) \] Thus, the new force required to stop the car moving with speed \( 3v \) is: \[ F_{{new}} = 9F \] Hence, the force needed to stop the car moving with speed \( 3v \) within the same distance is \( 9F \). \[ \boxed{9F} \]