NEET2014PhysicsWork, Power and EnergyActual
A spring of spring constant 5 10^3 ~N ~m ⁻¹ is stretched initially by 5 ~cm from the unstretched position. The work required to stretch it further by another 5 ~cm is
Options
- A6.25 ~N ~m
- B1250 ~N ~m
- C18.75 ~N ~m
- D25.00 ~N ~m
Correct answer
C. 18.75 ~N ~m
Step-by-step solution
U₁= 1 2 k x₁^2= 1 2 (5 10^3 ) (5 10⁻² )^2=6.25 ~N ~m U₂= 1 2 k x₂^2= 1 2 5 10^3 (5+5)^2 10⁻⁴=25 ~N ~m Work done =U₂-U₁=25.0-6.25=18.75 ~N ~m