Question

A block of mass $ 50\, \text{kg} $ is pulled at a constant speed of $ 4\, \text{m/s} $ across a horizontal floor by an applied force of $ 500\, \text{N} $ directed $ 30^\circ $ above the horizontal. The rate at which the force does work on the block in watts is:

• A. \( \dfrac{2000}{\sqrt{3}} \)
• B. \( 500\sqrt{3} \)
• C. 1732
• D. 1864

Hint:

Always resolve the force in the direction of motion to find power: \( P = F \cos \theta \cdot v \)

Answer 1

The Correct Option is C

The rate of work done by the applied force is given by: \[ P = F \cos \theta \cdot v \] Substitute values: \[ F = 500\, \text{N}, \quad \theta = 30^\circ, \quad v = 4\, \text{m/s} \] \[ P = 500 \cdot \cos 30^\circ \cdot 4 = 500 \cdot \frac{\sqrt{3}}{2} \cdot 4 = 1000\sqrt{3} \] \[ \Rightarrow P \approx 1000 \times 1.732 = 1732\, \text{watts} \]