A force F=(10+0.5x) acts on a particle in the x direction, where F is in newton and x is inmetre. Find the work done by this force during the displacement from x=0 to x=2.0m
A
18J
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B
19J
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C
20J
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D
21J
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Solution
The correct option is D21J The force acting on the body is variable, therefore we will integrate the work done by force during small displacement dx to obtain the net work done during the displacement from x=0 to x=2m.
Given, F=(10+0.5x)
The work done during small displacement (dx) is, dW=Fdx
Considering the force to be constant during small displacement dx ⇒W=∫20(10+0.5x)dx Or,W=[10x+14x2]20 ∴W=(20+1)−0=21J