Question

The force constant of a spring is 400N/m. How much work must be done on the

spring to stretch it (a) by 6.0cm (b) from x = 4.0cm to x = 6.0 cm, where x = 0 is the

relaxed position of the spring.

Answer 1

a). W=½*kx²
k=400N
x=0.06m
x²=0.0036m
W= ½×400×0.0036
=0.72J
b). W=½×kx²
∆x=6-4=2cm=0.02
W=½×400×0.0004
W=0.08J

Answer 2

[a]
W = 0.5kx²
= 0.5 × 400 N/m × (0.06 m)²
= 0.72 joules
Work done is 0.72 joules

[b]
W = 0.5k × (x₂² - x₁²)
= 0.5 × 400 N/m × [(0.06 m)² - (0.04 m)²]
= 0.4 joules
Work done is 0.4 joules