JEE Main Physics question
A spring of spring constant 5 × 10³ 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 [AIEEE 2003]
Answer: (C)
Solution
W = ½k(x₂² − x₁²) = ½(5×10³)[(0.10)² − (0.05)²] = ½(5×10³)(0.01 − 0.0025) = ½(5×10³)(0.0075) = 18.75 N-m.