Question

A particle of mass \( m \) at rest on a rough horizontal surface with a coefficient of friction \( \mu \) is given a velocity \( u \). The average power imparted by friction before it stops is:

• A. Zero
• B. \( \frac{1}{2} \mu m g u \)
• C. \( \mu m g u \)
• D. \( 2 \mu m g u \)

Hint:

To calculate average power due to friction, use the work-energy theorem and the time taken to stop, and then use the formula \( P = \frac{W}{t} \).

Answer 1

The Correct Option is B

The work done by friction is the force of friction times the distance the object travels. The force of friction is given by: \[ F_{\text{friction}} = \mu m g. \] The work done by friction is: \[ W = F_{\text{friction}} \times d = \mu m g \times d. \] Using the work
-energy theorem, we know that the work done by friction is equal to the change in kinetic energy. The initial kinetic energy is \( \frac{1}{2} m u^2 \), and the final kinetic energy is zero because the particle comes to rest. Thus: \[ \frac{1}{2} m u^2 = \mu m g \times d. \] The distance \( d \) can be found from the equation \( d = \frac{u^2}{2 \mu g} \). Now, the average power imparted by friction is the work done per unit time. The time taken to stop is: \[ t = \frac{u}{\mu g}. \] Thus, the average power is: \[ P = \frac{W}{t} = \frac{\mu m g \times \frac{u^2}{2 \mu g}}{\frac{u}{\mu g}} = \frac{1}{2} \mu m g u. \] Thus, the correct answer is \( \frac{1}{2} \mu m g u \).