JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 22)
A uniform disc of mass $0.5 \mathrm{~kg}$ and radius $r$ is projected with velocity $18 \mathrm{~m} / \mathrm{s}$ at $\mathrm{t}=0$ 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 __________ $\mathrm{J}$ (given, coefficient of friction is $0.3$ and $g=10 \mathrm{~m} / \mathrm{s}^{2}$ ).
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Answer
54
Explanation
$$v = {v_0} - \mu gt$$
$$ \Rightarrow v = 18 - 0.3 \times 10 \times 2 = 12$$ m/s
$$\Rightarrow$$ Kinetic energy $$ = {1 \over 2}m{v^2} + {1 \over 2}{{m{v^2}} \over 2}$$
$$ = {3 \over 4}m{v^2} = {3 \over 4} \times 0.5 \times 144\,\mathrm{J} = 54\,\mathrm{J}$$
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