InterviewSolution
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InterviewSolution
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A force `F=(10+0.50x)` acts on a particle in the x direction, where F is in newton and x in meter./find the work done by this force during a displacement form x=0 tox=2.0mA. 31.5 JB. 63 JC. 21 JD. 42 J |
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Answer» Correct Answer - C (c ) Given, force, `F=10+0.5x=10+(1)/(2)x` Let during small displacement the workdone by the force is dW=Fdx So, work done during displacement from x=0 to x=2 is `W=int_(0)^(w)dW=int_(0)^(2)Fdx=int_(0)^(2)(10+(1)/(2)x)dx` `=10[x]_(0)^(2)+[(x^(2))/(4)]_(0)^(2)=21J` |
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