NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
A block of mass m = 0 .1 kg is released from a height of 4 m on a curved smooth surface. On the horizontal surface, path AB is smooth and path BC is rough with a coefficient of friction μ = 0.1 . If the impact of the block with the vertical wall at C is perfectly elastic, the total distance covered by the block on the horizontal surface before coming to rest will be (take g = 10 m s - 2 )
Options
- A29   m
- B49   m
- C59 m
- D109   m
Correct answer
C. 59 m
Step-by-step solution
Total KE dissipated against friction = mgh = 4   J Work done against friction in one trip across BC = mgμ × 2 = 0 .2   J ∴ Total number of trips across BC = 4 0.2 = 20 ∴ Total distance = 20 × BC + 19 × AB = 59   m