MHT CET · Physics · Work Power Energy
A gardener pushes a lawn roller through a distance \(20 \mathrm{~m}\). If he applies a force of \(30 \mathrm{~kg}-\mathrm{wt}\) in a direction inclined at \(60^{\circ}\) to the ground, the work done by the gardener in pushing the roller is
\([\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}, \cos 30^{\circ}=\) \(\sin 60^{\circ}=\frac{\sqrt{3}}{2}]\)
- A \(3640 \mathrm{~J}\)
- B \(2460 \mathrm{~J}\)
- C \(3940 \mathrm{~J}\)
- D \(2940 \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(2940 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Consider the diagram below:

\(\mathrm{W}=\mathrm{F} \Delta \mathrm{S} \cos (\theta)\)
\(\mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\Delta \mathrm{S}}=\mathrm{F} \Delta \mathrm{S} \cos 60^{\circ}=30 \times(9.8) 20 \times \frac{1}{2}=2940 \mathrm{~J}\)

\(\mathrm{W}=\mathrm{F} \Delta \mathrm{S} \cos (\theta)\)
\(\mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\Delta \mathrm{S}}=\mathrm{F} \Delta \mathrm{S} \cos 60^{\circ}=30 \times(9.8) 20 \times \frac{1}{2}=2940 \mathrm{~J}\)
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