JEE Main2002PhysicsWork, Power and EnergyActual
A spring of force constant 800 ~N / m has an extension of 5 ~cm . The work done is extending it from 5 ~cm to 15 ~cm is
Options
- A16 ~J option 1 goes here
- B8 ~J
- C32 ~J
- D24 ~J
Correct answer
B. 8 ~J
Step-by-step solution
W = _ x₁ ^ x₂ F d x= _ x₁ ^ x₂ K x d x=K [ x^2 2 ]_ x₁ ^ x₂ = K 2 [x₂^2-x₁^2 ]= 800 2 [(0.15)^2-(0.05)^2 ]=8 ~J