Question:easy

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: Use the height-based definition of work instead of the force-component formula.
Work done by gravity depends only on the vertical height risen or fallen, not on the path taken: \[ W_{gravity} = -mgh \] where $h$ is the vertical rise. The minus sign shows gravity opposes upward motion.
Step 2: Find the vertical height gained while moving 4 m up the incline.
For an incline at angle $30^{\circ}$, moving a distance $d$ along the slope raises the block by \[ h = d \sin 30^{\circ} \] With $d = 4$ m: \[ h = 4 \times 0.5 = 2 \ \text{m} \]
Step 3: Plug in the numbers.
\[ W_{gravity} = -mgh = -(3)(10)(2) = -60 \ \text{J} \]
Step 4: Interpret the sign and the magnitude asked for.
The negative sign only tells us gravity acts opposite to the block's displacement, resisting the pull up the slope. The size of the work done is $60$ J.
Step 5: Cross-check against the setup.
The rope's tension supplies the extra pull that overcomes this $60$ J loss so the block keeps moving at constant speed, but the question only asks for the work done by the weight, which we found directly from the height risen, without ever needing the force component along the plane.
Final Answer:
\[ \boxed{60 \ \text{J}} \]
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