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 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:
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Always resolve the force in the direction of motion to find power: \( P = F \cos \theta \cdot v \)
Updated On: May 20, 2025
- \( \dfrac{2000}{\sqrt{3}} \)
- \( 500\sqrt{3} \)
- 1732
- 1864
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The Correct Option is C
Solution and Explanation
The rate of work done by the applied force is given by:
\[
P = F \cos \theta \cdot v
\]
Substitute values:
\[
F = 500\, \text{N}, \quad \theta = 30^\circ, \quad v = 4\, \text{m/s}
\]
\[
P = 500 \cdot \cos 30^\circ \cdot 4 = 500 \cdot \frac{\sqrt{3}}{2} \cdot 4 = 1000\sqrt{3}
\]
\[
\Rightarrow P \approx 1000 \times 1.732 = 1732\, \text{watts}
\]
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