The figure shows an 8.0 kg stone at rest on a spring. The spring is compressed 9

Question

The figure shows an 8.0 kg stone at rest on a spring. The spring is compressed 9.2 cm by the stone, (a) What is the spring constant? (b) The stone is pushed down an additional 35 cm and released. What is the elastic potential energy of the compressed spring just before that release? (c) What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height? (d) What is that maximum height, measured from the release point? (a) Number Units (b) Number Units (c) Number Units (d) Number Units

Explanation / Answer

a. Spring constant , K=F/X

Force, F= m*g

X is the displacement caused by the string

Gravitational field strength on Earth, g = 9.8 N/kg

So, F = m* g

=8* 9.8

=78.4 N

Displacement, x= 9.2 cm

= 9.2/100=0.092 m

Hence K = 78.4/0.092

= 852.17 N/m

b. Elastic potential energy, = 1/2 * k*x^2

Where k is spring constant

x is additional displacement

E.P.E=1/2*852.17*0.35^2

=52.2 J

C. Change in gravitational energy is same as elastic potential energy since the stone is released from rest position.

Change in potential energy = 52.2 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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