JEE Main Physics question
A spring of force constant 800 N m⁻¹ 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)(0.15² − 0.05²) = 400 × 0.02 = 8 J.
A spring of force constant 800 N m⁻¹ has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is
W = ½k(x₂²−x₁²) = ½(800)(0.15² − 0.05²) = 400 × 0.02 = 8 J.