A Spring Of Force Constant 800 Nm 1 Has An Extension Of 5cm

Q. A spring of force constant 800 Nm^-1 has an extension of 5cm. The work done in extending it from 5 cm to 15 cm is

1) 16 J

2) 8 J

3) 32 J

4) 24 J

Answer:

2) 8 J

Solution:

Force constant, k = 800 N/m

Work done in extending from 5 cm to 15 cm is

W = (1/2)k(x₂² – x₁²)

= (1/2) x 800 x (0.15² – 0.05²)

= (1/2) x 800 x (0.0225 – 0.0025)

= 400 x 0.02

= 8 N/m = 8J