An stone of mass M sits at rest on a spring near the surface of the Earth. If th

Question

An stone of mass M sits at rest on a spring near the surface of the Earth. If the spring is compressed a distance, d, by the weight of the stone, what is the spring constant k?

I set up an equation that says potential energy due to gravity initially equals the potential energy of the spring finally:

PE(gi) = PE (sf)
(m)(g)(d) = (0.5)(k)(d)^2
k = 2(m)(g)/(d)

Why is this incorrect? I found another answer that used the force of the spring to get the answer to be (m)(g)/(d), which made sense, but what about the previous method isn't correct?

Thanks.

Explanation / Answer

Gravitational energy mgd. Spring potential energy is 0.5kd^2 The d for the gravitational potential energy is not the same d in the spring potential energy. The d in gravitational potential energy is the distance from a fixed point point usually the the ground. The d in spring potential energy is the distance of compression. Therefore mgd does not equal 0.5kd^2 Instead use the force method Set up an equation were the gravitational force is equivalent to the force exerted by the spring mg=kd k=mg/d

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.