Question

45 POINTS

Answer 1

The elastic potential energy in the spring is 1 J

Explanation:

The elastic potential energy stored in a spring is given by:

$U=\frac{1}{2}kx^2$

where

k is the spring constant

x is the stretching/compression of the spring with respect to the unstretched length

For the spring in this problem, we have:

k = 200 N/m is its spring constant

The spring is stretched by 5.0 cm first and then by an additional 5.0 cm, so the total stretching is

$x=5.0 + 5.0 = 10.0 cm = 0.10 m$

Therefore, the elastic potential energy in the spring is:

$U=\frac{1}{2}(200)(0.10)^2=1 J$

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