Question

By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 7.00 cm to 15.00 cm? a) 159% b) 259% c) 359% d) 459%?

Answer 1

c) 359%

Step-by-step explanation:

k = spring constant of the spring

x₁ = initial stretch of the spring = 7 cm = 0.07 m

x₂ = final stretch of the spring = 15 cm = 0.15 m

Percentage increase in energy is given as

`(\Delta U * 100)/(U_(1))=\frac{((0.5)k{x_(2)}^(2) - {(0.5)k x_(1)}^(2))(100)}{(0.5)k{x_(1)}^(2)}`

`(\Delta U * 100)/(U_(1))=\frac{({x_(2)}^(2) - {x_(1)}^(2))(100)}{{x_(1)}^(2)}`

`(\Delta U * 100)/(U_(1))=\frac{({0.15}^(2) - {0.07}^(2))(100)}{{0.07}^(2)}`

`(\Delta U * 100)/(U_(1))` = 359%