A uniform disc of mass m and radius R starts with velocity $V_{0}$ on rough horizontal floor with a pure translational motion at t=0 as shown. At ${t=t}_{0}$ , disc starts rolling without sliding. Then (Coefficient of frictions is $μ$ )

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Net work done by frictional force upto time $t \leq t_{0}$ is given by $\frac{mμgt}{2} ({3μgt-2V}_{0})$

Net work done by frictional force upto time $t \leq t_{0}$ is given by $\frac{mμgt}{2} ({2μgt-V}_{0})$

Net work done by frictional force upto time ${t=2t}_{0}$ is given by ${mμgt}_{0} ({2μgt}_{0} {-V}_{0})$

Net work done by frictional force upto time ${t=2t}_{0}$ is given by $\frac{{mμgt}_{0}}{2} ({3μgt}_{0} {-2V}_{0})$

answer is A, D.

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Detailed Solution

$f = \frac{m r α}{2} f = μ m g α = \frac{2 μ g}{r} α_{c m} = μ g$
Displacement of the bottom most point $x = v_{0} t - \frac{1}{2} μ g t^{2} - \frac{1}{2} r α t^{2}$
Hence work done by friction $= W_{f} = - (μ m g x) = - μ m g (v_{0} t - \frac{3}{2} μ g t^{2})$
Once pure rolling begins work done by friction is zero.
Total work done at time $2 t_{0}$ is equal to work done till $t_{0}$ $W_{f} = - μ m g (v_{0} t_{0} - \frac{3}{2} μ g t_{0}^{2})$
Work done by friction while pure rolling is zero.