NTA Abhyas JEE Main2020PhysicsOscillationsPractice
The spring block system lies on a smooth horizontal surface. The free end of the spring is being pulled towards the right with constant speed V 0 = 2 m s - 1 . At t = 0 s , the spring of constant k = 100 N c m - 1 is unstretched and the block has a speed 1 m s - 1 to left. The maximum extension of the spring will be
Options
- A2 c m
- B4 c m
- C6 c m
- D8 c m
Correct answer
C. 6 c m
Step-by-step solution
In the frame (inertial w.r.t. earth) of the free end of spring, the initial velocity of the block is 3 m / s to left and the spring is unstretched. Applying conservation of energy between initial and maximum extension state, 1 2 m V 2 = 1 2 k A 2 A = m k V = 4 10000 × 3 = 6 c m