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=2.0m |
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Answer» Here, `F=(10+0.50x)` Small amount of word done in moving the particle through a samll distance dx is `dW=hat(F).d hatx=(10+0.5x)dx` Total work done, `W=int_(x=0)^(x=2)(10+00.5x)dx` `W=[10x +0.5(x^(2))/(2)]_(0)^(2)` `=10(2-0)+(0.5)/(2)(2^(2)-0)=20+1` `=21joule` |
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