JEE MAIN - Physics (2023 - 10th April Morning Shift - No. 23)
A closed circular tube of average radius 15 cm, whose inner walls are rough, is kept in vertical plane. A block of mass 1 kg just fit inside the tube. The speed of block is 22 m/s, when it is introduced at the top of tube. After completing five oscillations, the block stops at the bottom region of tube. The work done by the tube on the block is __________ J. (Given g = 10 m/s$$^2$$).
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Answer
245
Explanation
From work energy theorem
$$ \begin{aligned} &W_{\text {gravity }}+W_{\text {friction }} =\Delta(K E)=K E_f-K E_i \\\\ &W_{\text {gravity }} =m g h=1 \times 10 \times 0.3=3 \mathrm{~J} \\\\ &W_{\text {friction }} =0-\frac{1}{2} \times(22)^2-3 \\\\ & =-(242+3)=-245 \mathrm{~J} \end{aligned} $$
$$ \begin{aligned} &W_{\text {gravity }}+W_{\text {friction }} =\Delta(K E)=K E_f-K E_i \\\\ &W_{\text {gravity }} =m g h=1 \times 10 \times 0.3=3 \mathrm{~J} \\\\ &W_{\text {friction }} =0-\frac{1}{2} \times(22)^2-3 \\\\ & =-(242+3)=-245 \mathrm{~J} \end{aligned} $$
The negative sign indicates that the work done by frictional force (the tube) is in the direction opposite to the displacement of the block. In conclusion, the work done by the tube on the block is 245 J.
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