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