Work Energy Power

JEE Main level, numerical answer, medium

67 of 183

Question 67 of 183

A closed circular tube of average radius 15cm15cm, whose inner walls are rough, is kept in vertical plane. A block of mass 1kg1kg just fit inside the tube. The speed of block is 22ms−122m{s}^{-1}, 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 ________ JJ. (Given g=10ms−2g=10m{s}^{-2}).

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Answer: 245

The data given is
r=15cmm=1kgv=22ms−1r=15cm m=1kg v=22m{s}^{-1}
Total work done by the tube on the block is equal to loss in mechanical energy,
w=12mv2+mghw=\dfrac{1}{2}m{v}^{2}+mgh
⇒w=12×1×(22)2+1×10×0.3\Rightarrow w=\dfrac{1}{2}\times 1\times {\left(22\right)}^{2}+1\times 10\times 0.3
⇒w=245J\Rightarrow w=245J

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