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

Option: 1

54


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{a}=-\mu_{\mathrm{k}} \mathrm{g}=-3
\mathrm{v}=\mathrm{u}+\mathrm{at}
\mathrm{v}=18-3 \times 2=12 \mathrm{~ms}^{-1}
\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{\mathrm{mr}^{2}}{2}\right)\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^{2}
\mathrm{KE}=\frac{3}{4} \mathrm{mv}^{2}
\mathrm{KE}=3 \times 18=54 \mathrm{~J}

 

Posted by

Pankaj Sanodiya

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