Question

A block of mass 20 kg is moved with constant velocity along an inclined plane of inclination 37° with help of a force of constant power 50 W. If the coefficient of kinetic friction between block and surface is 0.25, then what fraction of power is used against gravity?

• A. $\frac{3}{4}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{8}$

Answer 1

Answer: (A)

$\frac{3}{4}$

Answer 2

The power used to overcome gravity is given by:

$$P_{\text{gravity}} = mg \sin\theta \times v$$

• The total power output must overcome both gravity and friction:

$$P = mg(\sin\theta + \mu \cos\theta) \times v$$

• The fraction of power used against gravity is:

$$\text{Fraction} = \frac{mg \sin \theta \, v}{mg(\sin\theta + \mu \cos\theta) \, v} = \frac{\sin\theta}{\sin\theta + \mu \cos\theta}.$$

• With $\theta = 37^\circ$ ($\sin 37^\circ \approx 0.6,\; \cos 37^\circ \approx 0.8$) and $\mu = 0.25$:

$$\text{Fraction} = \frac{0.6}{0.6 + 0.25 \times 0.8} = \frac{0.6}{0.6 + 0.2} = \frac{0.6}{0.8} = 0.75 = \frac{3}{4}.$$