Question

A motor rotates a pulley of radius 25 cm at 20 rpm. A rope around the pulley lifts a 50 kg block, the power output of the motor is

• A. 261 w
• B. 216 w
• C. 305 w
• D. 117 w

Answer 1

Answer: (A)

261 w

Answer 2

The problem asks for the power output of a motor that lifts a block using a pulley.

Here's a step-by-step solution:

1. Identify the given parameters:

   • Radius of the pulley, $R = 25 \text{ cm} = 0.25 \text{ m}$
   • Rotational speed of the pulley, $N = 20 \text{ rpm}$ (revolutions per minute)
   • Mass of the block, $m = 50 \text{ kg}$

2. Convert rotational speed to angular velocity ($\omega$): The rotational speed $N$ in rpm can be converted to angular velocity $\omega$ in radians per second using the formula: $\omega = N \times \frac{2\pi \text{ rad}}{1 \text{ revolution}} \times \frac{1 \text{ minute}}{60 \text{ seconds}}$ $\omega = 20 \times \frac{2\pi}{60} \text{ rad/s} = \frac{40\pi}{60} \text{ rad/s} = \frac{2\pi}{3} \text{ rad/s}$

3. Calculate the force required to lift the block: The force required to lift the block at a constant velocity is equal to its weight. We'll use $g = 10 \text{ m/s}^2$ as it often leads to options in competitive exams. $F = mg = 50 \text{ kg} \times 10 \text{ m/s}^2 = 500 \text{ N}$

4. Calculate the linear speed ($v$) of the rope: The linear speed of the rope is the tangential speed of the pulley's rim: $v = R\omega$ $v = 0.25 \text{ m} \times \frac{2\pi}{3} \text{ rad/s} = \frac{0.5\pi}{3} \text{ m/s} = \frac{\pi}{6} \text{ m/s}$

5. Calculate the power output ($P$) of the motor: The power output of the motor is the product of the force applied and the linear speed: $P = F \cdot v$ $P = 500 \text{ N} \times \frac{\pi}{6} \text{ m/s} = \frac{500\pi}{6} \text{ W} = \frac{250\pi}{3} \text{ W}$

   Now, substitute the value of $\pi \approx 3.14159$: $P = \frac{250 \times 3.14159}{3} = \frac{785.3975}{3} \approx 261.799 \text{ W}$

   Rounding this value, we get approximately $262 \text{ W}$. Among the given options, $261 \text{ W}$ is the closest.