Question

How much energy is in the elastic potential energy store of a spring with a spring constant of 4 N/m if it is stretched from 1 m to 1.1 m in length?

Answer 1

0.02 J

Step-by-step explanation:

In order to find the energy in the elastic potential energy store of the spring we use the formula

`U = (1)/(2) k Δ {x}^(2) \\`

where

U is the elastic potential energy in Joules

k is the spring constant in N/m

Δx is the deformation(stretch or compression) of the spring in metres

from the question

k = 4 N/m

Δx = 1.1 m - 1 m = 0.1 m

We have

`U = (1)/(2) * 4 * ( {0.1})^(2) \\ = 2 * 0.01 \\ = 0.02`

We have the final answer as

0.02 J

Hope this helps you