InterviewSolution
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InterviewSolution
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The kinetic energy K of a particle moving along a circle of radius R depends upon the distance s as `K=as^2`. The force acting on the particle isA. (a) `2a(s^2)/(R)`B. (b) `2as[1+(s^2)/(R)]^(1//2)`C. (c) `2as`D. (d) `2a` |
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Answer» Correct Answer - B Given that `K=as^2` or `1/2mv^2=as^2` or `mv^2=2as^2` (i) Differenciating w.r.t. time, we get `m_2vxx(dv)/(dt)=2axx2sxx(ds)/(dt)` But `(ds)/(dt)=v` So `2m(dv)/(dt)=4as` or `m(dv)/(dt)=2as` Now, `m(dv)/(dt)`=tangential force `=F_t` `F_t=2as` Centripetal force `=F_r(mv^2)/(R)=(2as^2)/(R)` `F_(n et)=sqrt(F_t^2+F_r^2)=sqrt((2as)^2+((2as^2)/(R))^2)` `=2assqrt(1+s^2/R^2)` |
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