AP EAMCET202418 May 2024Morning ShiftPhysicsLaws of MotionActual
A spring of 5 10^3 Nm ⁻¹ spring constant is stretched initially by 10 cm from unstretched position. The work required to stretch it further by another 10 cm is
Options
- A75 ~N - m
- B50 ~N - m
- C76 ~N - m
- D82 ~N - m
Correct answer
A. 75 ~N - m
Step-by-step solution
The work done required to stretch a spring from x ₁ to x₂ is W = 1 2 k ( x ₂^2- x ₁^2 )= 1 2 5 10^3 [ (20 10⁻² )^2- (10 10⁻² )^2 ]=75 ~N - m