JEE Main Physics question
A spring of spring constant 5 × 10³ N m⁻¹ is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is
Answer: (B)
Solution
(Same as Q1.) W = ½k(x₂²−x₁²) = ½(5000)(0.10²−0.05²) = 18.75 J.