A certain 2.00 kg object is hung vertically on a certain light asdescribed by ho

Question

A certain 2.00 kg object is hung vertically on a certain light asdescribed by hooke's law, the spring stretches to 3.30 cm.
What is the force constant of the spring n/m?
If the 2.00 kg object is moved how far will the springstretch? (cm)
How much work must be done to stretch the same 8.70 cm fromits unstretched position?


What is the force constant of the spring n/m?
If the 2.00 kg object is moved how far will the springstretch? (cm)
How much work must be done to stretch the same 8.70 cm fromits unstretched position?

Explanation / Answer

Given that x = 3.30 cm                   m = 2.00 kg so that we have    F = mg = 2 * 9.8     = 19.6 N The force constant of the spring k = F / x = 19.6 N / 0.033 m     = 593.93 N/m Workdone to stretch the same 8.70 cm from its unstretchedposition is            W = F.x = kx2    = 593.93 * 0.0872                                      = 4.49 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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