A uniform disc of mass \(0.5\,kg\) and radius \(r\) is projected with velocity \(18\,m / s\) at \(t =0\,s\) on a rough horizontal surface. It starts off with a purely sliding motion at \(t =0\,s\). After \(2\,s\) it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after \(2\,s\) will be \(..............J\) (given, coefficient of friction is \(0.3\) and \(\left.g =10\,m / s ^2\right)\)

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Answer & Solution

Correct Answer

(C) \(54\)

Step-by-step Solution

Detailed explanation

\(a =-\mu_{ k } g =-3\)

\(V =18-3 \times 2\)

\(V =12\,m / s\)

\(KE =\frac{1}{2} mv ^2+\frac{1}{2} \frac{ mr ^2}{2} \frac{ v ^2}{ r ^2}\)

\(KE =\frac{3}{4} mv ^2\)

\(KE =3 \times 18=54\,J\)

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