When a 2.20 kg object is hung vertically on a certain lightspring described by H

Question

When a 2.20 kg object is hung vertically on a certain lightspring described by Hooke's Law, the spring stretches.00327m. a)what is the force constant of the spring? b)if the 2.2 kg object removed, how far will the stringstretch with 1.10 kg block hung on it? c) how much work must an external agent do to stretch thesame spring .0076m from unstretched position? When a 2.20 kg object is hung vertically on a certain lightspring described by Hooke's Law, the spring stretches.00327m. a)what is the force constant of the spring? b)if the 2.2 kg object removed, how far will the stringstretch with 1.10 kg block hung on it? c) how much work must an external agent do to stretch thesame spring .0076m from unstretched position?

Explanation / Answer

a) F=-kx 2.2kg=-k(.00327m) k=672.78N/m b)x=F/-k x=1.10/672.78 x=.001635m=1.635cm c)?

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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