JEE Main Physics question
A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is
Answer: (B)
Solution
W = ½k(x₂² − x₁²) = ½(800)[(0.15)² − (0.05)²] = ½(800)(0.0225 − 0.0025) = ½(800)(0.02) = 8 J. ## Conservation of Energy and Momentum