Question:
A block of mass 3.0 kg is pulled at a constant speed with a taut rope along a frictionless plane that is inclined at 30°. Then the work done by the weight of the block if it is pulled a distance 4 m along the inclined plane in joule is [Acceleration due to gravity = 10 m s$^{-2}$]:
A block of mass 3.0 kg is pulled at a constant speed with a taut rope along a frictionless plane that is inclined at 30°. Then the work done by the weight of the block if it is pulled a distance 4 m along the inclined plane in joule is [Acceleration due to gravity = 10 m s$^{-2}$]:
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On an inclined plane, component of weight along the plane is \(mg \sin \theta\). Work done by weight is negative if displacement is upward.
Updated On: Jul 18, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the problem.
A block moves up a frictionless inclined plane with constant speed. We are asked to find the work done by gravity along the incline. Work done by a force is defined as: \[ W = F \cdot d \cdot \cos \theta \] where \(F\) is force, \(d\) is displacement, and \(\theta\) is the angle between force and displacement.
Step 2: Identify the force due to weight along the incline.
The component of gravitational force along the incline is: \[ F_{\text{gravity}} = mg \sin \alpha \] where \(m = 3\) kg, \(g = 10 \, \text{m/s}^2\), and \(\alpha = 30^\circ\).
Step 3: Compute the force component.
\[ F_{\text{gravity}} = 3 \times 10 \times \sin 30^\circ = 30 \times 0.5 = 15 \, \text{N} \]
Step 4: Apply the work formula.
The block is pulled 4 m along the plane. Since gravity acts downward along the incline (opposite to displacement), the work done by gravity is: \[ W = F \cdot d \cdot \cos 180^\circ = 15 \times 4 \times (-1) = -60 \, \text{J} \] Magnitude of work is 60 J.
Step 5: Sign interpretation.
Negative sign indicates work done by gravity is opposite to displacement (resisting motion). Magnitude is requested.
Step 6: Final conclusion.
\[ \boxed{60 \, \text{J}} \]
A block moves up a frictionless inclined plane with constant speed. We are asked to find the work done by gravity along the incline. Work done by a force is defined as: \[ W = F \cdot d \cdot \cos \theta \] where \(F\) is force, \(d\) is displacement, and \(\theta\) is the angle between force and displacement.
Step 2: Identify the force due to weight along the incline.
The component of gravitational force along the incline is: \[ F_{\text{gravity}} = mg \sin \alpha \] where \(m = 3\) kg, \(g = 10 \, \text{m/s}^2\), and \(\alpha = 30^\circ\).
Step 3: Compute the force component.
\[ F_{\text{gravity}} = 3 \times 10 \times \sin 30^\circ = 30 \times 0.5 = 15 \, \text{N} \]
Step 4: Apply the work formula.
The block is pulled 4 m along the plane. Since gravity acts downward along the incline (opposite to displacement), the work done by gravity is: \[ W = F \cdot d \cdot \cos 180^\circ = 15 \times 4 \times (-1) = -60 \, \text{J} \] Magnitude of work is 60 J.
Step 5: Sign interpretation.
Negative sign indicates work done by gravity is opposite to displacement (resisting motion). Magnitude is requested.
Step 6: Final conclusion.
\[ \boxed{60 \, \text{J}} \]
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