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A force acting on particle is given by `vecF=(3x^2hati+4yhatj)N`. The change in kinetic energy of particle as it moves from `(0,2m)` to `(1m,3m)` isA. `6J`B. `10J`C. `11J`D. `13J` |
Answer» Correct Answer - C `vecF=3x^2hati+yhatj`,`dvecr=dxhati+dyhatj` `dW=vecF.dvecr=3x^2dx+4ydy` `W=int_(0)^(1)3x^2dx+int_(2)^(3)4ydy` `=|x^3|_0^1+|2y^2|_2^3` `=1+2(3^2-2^2)=11J` `W=triangleK=11J` |
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