JEE Main2003PhysicsWork, Power and EnergyActual
A spring of spring constant 5 10^3 ~N / m is stretched initially by 5 ~cm from the unstretched position. Then the work required to stretch it further by another 5 ~cm is
Options
- A12.50 ~N - m
- B18.75 ~N - m
- C25.00 ~N - m
- D6.25 ~N - m
Correct answer
B. 18.75 ~N - m
Step-by-step solution
Required work done = 1 2 ~K ( x ₂^2- x ₁^2 )= 1 2 5 10^3 [10^2-5^2 ] 10⁻⁴= 1 2 5 75 10^3 10⁻⁴=18.75