Question:
A block of mass 2 kg is pulled at a constant speed with a taut rope along a frictionless plane that is inclined at \(30^\circ\). Find the work done by the tension in the rope in pulling it a distance 4 m along the inclined plane. (Acceleration due to gravity \(g = 10 \, \text{m/s}^2\))
A block of mass 2 kg is pulled at a constant speed with a taut rope along a frictionless plane that is inclined at \(30^\circ\). Find the work done by the tension in the rope in pulling it a distance 4 m along the inclined plane. (Acceleration due to gravity \(g = 10 \, \text{m/s}^2\))
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For constant speed on a frictionless incline, work done by tension equals component of weight along the incline multiplied by distance: \(W = mg \sin \theta \cdot d\).
Updated On: Jul 18, 2026
- 40 J
- 20 J
- 68 J
- 136 J
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The Correct Option is A
Solution and Explanation
Step 1: Analyze forces.
The block is pulled at constant speed, so net force along the plane is zero. Tension \(T\) balances the component of weight along the incline:
\[ T = mg \sin \theta \]
Step 2: Substitute values.
\[ T = 2 \times 10 \times \sin 30^\circ = 20 \times 0.5 = 10 \, \text{N} \]
Step 3: Calculate work done.
Work done by the tension along the plane over distance \(d = 4 \, \text{m}\) is:
\[ W = T \cdot d = 10 \times 4 = 40 \, \text{J} \]
Step 4: Verify reasoning.
Since the plane is frictionless and motion is at constant speed, no extra work is done; all work goes into counteracting the component of gravity.
Step 5: Final conclusion.
Hence, the work done by the tension in the rope is:
\[ \boxed{40 \, \text{J}} \]
The block is pulled at constant speed, so net force along the plane is zero. Tension \(T\) balances the component of weight along the incline:
\[ T = mg \sin \theta \]
Step 2: Substitute values.
\[ T = 2 \times 10 \times \sin 30^\circ = 20 \times 0.5 = 10 \, \text{N} \]
Step 3: Calculate work done.
Work done by the tension along the plane over distance \(d = 4 \, \text{m}\) is:
\[ W = T \cdot d = 10 \times 4 = 40 \, \text{J} \]
Step 4: Verify reasoning.
Since the plane is frictionless and motion is at constant speed, no extra work is done; all work goes into counteracting the component of gravity.
Step 5: Final conclusion.
Hence, the work done by the tension in the rope is:
\[ \boxed{40 \, \text{J}} \]
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