A force F acting on a particle varies with the position x as shown in figure. The work done by this force in displacing the particle from x=−2m to x=0.
A
+10J
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B
+20J
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C
−20J
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D
−10J
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Solution
The correct option is D−10J While moving x=−2m to x=0, the particle has a positive displacement but the force acting on the particle is along negative x direction. ∴ Work done by the force is negative.
Hence we can write for workdone by force as, W=−[Area under (F−x) graph] ∴W=−[12(2)(10)]=−10J