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The kinetic energy of a particle moving along a circle of radius `R` depends on the distance covered `s` as `K=lambdas^(2)`, where `lambda` is a constant. Find the force acting on the particle as a function of `s`. |
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Answer» `KE = (1)/(2) mv^(2) = cs^(2) rArr v = (sqrt((2c)/(m))) s` `a_(t) = (dv)/(dt) = sqrt((2c)/(m)) xx (ds)/(dt) = v sqrt((2c)/(m))` `F_(t) =ma_(t) = mv sqrt((2c)/(m)) = [m sqrt((2c)/(m))s] sqrt((2c)/(m)) = 2cs` Total force `F = sqrt(F_(t)^(2)+ F_(c)^(2))= sqrt((2cs)^(2) + ((mv^(2))/(r))^(2))` `F = 2cs sqrt(1+ (s^(2))/(r^(2)))`. |
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