1.

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`


Discussion

No Comment Found

Related InterviewSolutions