Question

A scooter moves with a speed of 7 m/s on a straight road and is stopped by applying the brakes. Before stopping, the scooter travels 10 m. If the weight of the scooter is W, then the total resistance to the motion of the scooter will be

• A. \(\frac{1}{2} W\)
• B. \(\frac{1}{4} W\)
• C. 4W
• D. 2W

Hint:

When solving for work and energy, ensure that units are consistent, especially when dealing with mass and weight.

Answer 1

The Correct Option is B

We will use the work-energy principle to solve this problem. The work done by the resistive force \( F_{\text{res}} \) is equal to the change in kinetic energy. The kinetic energy \( K \) is given by: \[ K = \frac{1}{2} m v^2 \] where \( m \) is the mass of the scooter and \( v \) is its speed. The initial kinetic energy of the scooter before stopping is: \[ K_i = \frac{1}{2} m v^2 \] The final kinetic energy \( K_f \) is zero because the scooter is stopped. The work done by the resistive force is given by: \[ W = F_{\text{res}} \cdot d \] where \( d \) is the distance over which the force acts (10 m). By the work-energy principle: \[ F_{\text{res}} \cdot d = \Delta K = K_i - K_f \] Substituting the values: \[ F_{\text{res}} \cdot 10 = \frac{1}{2} m v^2 \] We know the weight \( W = mg \), so \( m = \frac{W}{g} \). Substitute this into the equation: \[ F_{\text{res}} \cdot 10 = \frac{1}{2} \frac{W}{g} v^2 \] Since \( v = 7 \, \text{m/s} \), substitute that in: \[ F_{\text{res}} \cdot 10 = \frac{1}{2} \frac{W}{g} (7)^2 \] Simplifying further: \[ F_{\text{res}} \cdot 10 = \frac{1}{2} \frac{W}{g} \cdot 49 \] \[ F_{\text{res}} = \frac{49W}{20g} \] Now, to find the total resistance to the motion, we note that the resistance \( F_{\text{res}} \) is proportional to the weight of the scooter \( W \).
Thus, the total resistance \( R \) to the motion of the scooter is: \[ R = \frac{1}{4} W \]