NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
A vertical spring of force constant 100 Nm - 1 is attached to the ceiling, with a block of mass 10 kg suspended from it. Now an external force F is applied on the block so that the spring is stretched by an additional length of 2 m . The work done by the force F is [Take g = 10 ms - 2 ]
Options
- A200   J
- B400   J
- C100   J
- D600   J
Correct answer
A. 200   J
Step-by-step solution
x = elongation in spring due to mass 10 kg = 1 0 × 1 0 1 0 0 = 1 m W F = 1 2 × 1 0 0 × 3 2 - 1 2 - 1 0 × 1 0 × 2 = 200   J