Kerala KEAM 5 June 2024 Solved Paper

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 $9F$.$$\boxed{9F}$$ Quick Tip: 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 $3^2 = 9$.