When a 2.00-kg object is hungvertically on a certain light spring described by H

Question

When a 2.00-kg object is hungvertically on a certain light spring described by Hooke's law, thespring stretches 3.30cm. (a) What is the force constant of the spring?
N/m

(b) If the 2.00-kg object isremoved, how far will the spring stretch if a 1.00-kg block is hung on it?
cm

(c) How much work must an external agent do to stretch the samespring 9.40 cm from itsunstretched position?
J (a) What is the force constant of the spring?
N/m

(b) If the 2.00-kg object isremoved, how far will the spring stretch if a 1.00-kg block is hung on it?
cm

(c) How much work must an external agent do to stretch the samespring 9.40 cm from itsunstretched position?
J

Explanation / Answer

a) Spring constant =Force per unitextension = F / x = 2*9.8 /0.0330 = 593.94 N/m

b) 1 * 9.8 = 593.94 x

x= 0.0165 m = 1.65 cm

c) Work = kx2/2

w = 593.94 * 0.094 * 0.094 / 2 =2.62 J

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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