Question:
A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1 m. Speed of the block at the bottom is 2 m/s. Work done by the frictional force on the block when it reaches at the bottom is
A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1 m. Speed of the block at the bottom is 2 m/s. Work done by the frictional force on the block when it reaches at the bottom is
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Work done by non-conservative forces = change in mechanical energy.
Updated On: Apr 16, 2026
- 8 J
- -8 J
- 4 J
- 0 J
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Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Initial PE = $mgh = 1 × 10 × 1 = 10$ J.
Step 2: Final KE = $\frac12mv² = \frac12 × 1 × 4 = 2$ J.
Step 3: Work done by friction = change in mechanical energy = $2 - 10 = -8$ J.
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