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 \(22 \mathrm{~m} / \mathrm{s}\), 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 \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) ]

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Work, Power & Energy

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}

]

Answer: -245

Solution

By the work-energy theorem the total work equals the change in kinetic energy:

W_{tube} + W_{gravity} = Δ K E

. The block enters at the top with

K E = \frac{1}{2} (1) (22)^{2} = 242

J and finally rests, so

Δ K E = - 242

J. Travelling from the top to the bottom of the tube lowers it by one diameter,

2 r = 0.30

m, so gravity does

W_{gravity} = m g (2 r) = 1 \times 10 \times 0.30 = 3

J. Hence the work done by the tube is

W_{tube} = Δ K E - W_{gravity} = - 242 - 3 = - 245

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Integer answer

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JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Work, Power & Energy

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