Question

A spring with a force constant of 351

N/m is in a compressed state, 41. 5
cm shorter than its relaxed length. In this condition, the spring serves a vital purpose in a device for detecting impending earthquakes. A curious physicist releases the spring and, moreover, stretches it until it is 15. 1
cm longer than its relaxed length. This renders the earthquake detector useless, and it fails to warn of a very strong earthquake that subsequently shatters all the lawn sculptures in the area, with no other damage or injury. Although some might consider this to be fortunate, the physicist's boss does not. She had some extremely unattractive plaster gnomes around her yard and loved them dearly. As a result, she imposes a punishment on the physicist and requires him to calculate the change of elastic potential energy caused by his meddling. What does he find?

Answer 1

The change in elastic potential energy caused by stretching or compressing a spring with a known force constant can be calculated using the formula:

ΔPE = 1/2 kΔx^2

Where:

- ΔPE is the change in elastic potential energy

- k is the force constant of the spring (in N/m)

- Δx is the change in the length of the spring (in meters)

In this case, the spring is initially compressed by 41.5 cm shorter than its relaxed length, which is equivalent to a displacement of -0.415 m. When the spring is stretched, it is extended by 15.1 cm longer than its relaxed length, which is equivalent to a displacement of 0.151 m.

So, the change in elastic potential energy is given by:

ΔPE = 1/2 k (0.151 - (-0.415))^2

ΔPE = 1/2 (351 N/m) (0.151 + 0.415)^2

ΔPE ≈ 45.5 joules

Therefore, the change in elastic potential energy caused by the physicist's meddling is approximately 45.5 joules.