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A particle of mass m is subjected to a force `vecF=F_(0)[cos (t)hati+sin(1)hatj]`. If initially (t=0) the particle was at rest, the kinetic energy of the particle as a function of time is given by:A. `(F_(0)^(2))/(m)(1-cos(2t))`B. `(F_(0)^(2))/(m)(1-cost)`C. `(F_(0)^(2))/(m) sin(t)`D. `(F_(0)^(2))/(m)t` |
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Answer» `veca=(vecF)/(m)=(F_(0))/(m) [cos(t)hati+sin(t)hatj]` `vecv=int_(0)^(t)vecadt=(F_(0))/(m)[sin(t)hati+(1-cos t)hatj]` `KE=(1)/(2)m(vecv.vecv)=(1)/(2)m[(F_(0)^(2))/(m^(2))(2-2cps t)]` `=(F_(0)^(2))/(m) (1-cos t)` |
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