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 ms−2)
Correct Answer: 245
Solution and Explanation
Step 1: Given data
The radius of the arc is given as:
\[ R_{\text{arc}} = 15 \, \text{cm} = 0.15 \, \text{m}. \]
Step 2: Work-energy theorem (WET)
By the work-energy theorem (WET), we have:
\[ W_f + W_{\text{gravity}} = \Delta K = K_f - K_i, \] where: - \( W_f \) is the work done by friction, - \( W_{\text{gravity}} \) is the work done by gravity, - \( K_f \) is the final kinetic energy, - \( K_i \) is the initial kinetic energy.
Step 3: Substitute the known values
The equation becomes:
\[ W_f + 10 \times 0.3 = 0 - \frac{1}{2} \times 1 \times (22)^2. \] Simplifying: \[ W_f + 3 = 0 - \frac{1}{2} \times 484. \]
Step 4: Solve for \( W_f \)
Now, simplify further:
\[ W_f + 3 = -242. \] Rearranging: \[ W_f = -245 \, \text{J}. \]
Step 5: Work by friction
The work done by friction is:
\[ W_f = -245 \, \text{J}. \]
Final Answer:
The work done by friction is:
\[ W_f = -245 \, \text{J}. \]
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