Force F on a particle moving in a straight line varies with distance d as shown in the figure The work done on the particle during its displacement of 12m is
A
18J
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B
21J
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C
26J
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D
13J
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Solution
The correct option is D13J Total work done= W=∫123→F.d→x also, Wtrap=12×h× sum of parallel sides W=12×(2−0)×(9+4)W=13J