Question

# 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=2.0 m

A
18 J
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B
19 J
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C
20 J
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D
21 J
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Solution

## The correct option is D 21 JThe 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=2 m. 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=21 J

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