Figure 10-34 shows an 8.00 kg stone resting on a spring. The spring is compresse

Question

Figure 10-34 shows an 8.00 kg stone resting on a spring. The spring is compressed 7.0 cm by the stone.

(a) What is the spring constant?
N/cm
(b) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?
J
(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?
J
(d) What is that maximum height, measured from the release point?

Explanation / Answer

a) kx = mg k = mg/x = 8*9.81/7E-2=1121 N/m =11.21 N/cm b) PE = 1/2 k x^2 =0.5*1121*(.37)^2=76.73 J c) all this snergy goes into graviation PE so 76.73 d) 76.73 = m g h h = 76.73/(8*9.81)=0.978 m

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

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