Question

Power of an engine driving a vehicle of mass \(m\) with a speed \(v\) on a horizontal road is (\(\mu\) is the coefficient of friction between the road and the tyre)

• A. \(\frac{mg}{\mu v}\)
• B. \(\mu mgv\)
• C. \(\mu mgv^2\)
• D. \(\frac{\mu mg}{v}\)
• E. \(\frac{\mu mv}{g}\)

Hint:

For motion at constant speed, engine power is used only to overcome resistive forces like friction: \(P=Fv\).

Answer 1

The Correct Option is B

Step 1: Identify the resisting force.
On a horizontal road, the resistive force due to friction is: \[ F=\mu N \] Since the normal reaction: \[ N=mg, \] we get: \[ F=\mu mg \]

Step 2: Recall the formula for power.
Power is given by: \[ P=\text{Force}\times \text{velocity} \]

Step 3: Substitute the force.
\[ P=\mu mg \cdot v \]

Step 4: Interpret the expression.
This represents the power required to overcome friction and keep the vehicle moving at constant speed.

Step 5: Check units.
\[ (\text{force})\times(\text{velocity})=\text{N}\cdot \text{m/s}=\text{W} \] So the expression is dimensionally correct.

Step 6: Compare with the options.
The expression \(\mu mgv\) matches option \((2)\).

Step 7: Final answer.
\[ \boxed{\mu mgv} \]