A force F=5+2x+3x2 acts on a particle in x direction where F is in Newton and x is in meter . Find the workdone by this force during a displacement from x=1m to x=2m.
A
15J
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B
12J
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C
32J
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D
−15J
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Solution
The correct option is A15J Given, variable force F=5+2x+3x2
Workdone by variable force (W)=∫x2x1FxdxW=∫21(5+2x+3x2)dxW=5x|21+2x22|21+3x33|21W=5(2−1)+(22−12)+(23−13)W=5+(4−1)+(8−1)=15J