Question:

A block of mass \(50\,\text{kg}\) is pulled at a constant speed of \(4\,\text{m s}^{-1}\) 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 watt is

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Power delivered by a force is \[ P=Fv\cos\theta \] where \(\theta\) is the angle between force and velocity.
Updated On: Jun 22, 2026
  • \(\frac{2000}{\sqrt{3}}\)
  • \(500\sqrt{3}\)
  • \(1732\)
  • \(1864\)
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The Correct Option is C

Solution and Explanation

Step 1: Use the formula for power.
The rate at which a force does work is called power.
Power is given by \[ P=\vec{F}\cdot \vec{v} \] or \[ P=Fv\cos\theta \] where \[ F=500\,\text{N} \] \[ v=4\,\text{m s}^{-1} \] and \[ \theta=30^\circ \]

Step 2: Substitute the values.
\[ P=500\times 4\times \cos30^\circ \] Since, \[ \cos30^\circ=\frac{\sqrt{3}}{2} \] Therefore, \[ P=500\times 4\times \frac{\sqrt{3}}{2} \] \[ P=1000\sqrt{3} \]

Step 3: Calculate the numerical value.
Using \[ \sqrt{3}\approx 1.732 \] \[ P\approx 1000\times 1.732 \] \[ P\approx 1732\,\text{W} \]

Step 4: Final conclusion.
Therefore, the rate at which the force does work is \[ \boxed{1732\,\text{W}} \]
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