Question

A particle of mass m moving from origin along x-axis and its velocity varies with position x as v= k√x. The work done by force acting on it during first 't' seconds is

Joshi sir comment

v = k√x

so a = dv/dt = [k/2√x]*dx/dt  = [k/2√x]*v = [k/2√x]*[k√x] = k2/2 = constant

and dx/dt = k√x 

so dx/√x = kdt

integrate within o to t

finally work = Fx  because F is constant

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