When a 2.50-kg object is hungvertically on a certain light spring described by H

Question

When a 2.50-kg object is hungvertically on a certain light spring described by Hooke's law, thespring stretches 3.31 cm. (a) What is the force constant of thespring?
1 N/m

(b) If the 2.50-kg object is removed,how far will the spring stretch if a 1.25-kg block is hung on it?
2 cm

(c) How much work must an external agent do to stretch the samespring 8.90 cm from its unstretchedposition?
3 J (a) What is the force constant of thespring?
1 N/m

(b) If the 2.50-kg object is removed,how far will the spring stretch if a 1.25-kg block is hung on it?
2 cm

(c) How much work must an external agent do to stretch the samespring 8.90 cm from its unstretchedposition?
3 J

Explanation / Answer

a) F=kx, so 2.5kg(9.8)=k(3.31x10-2m) and k=740.18N/m b) x=((1.25kg)(9.8))/(740.18N/m)= 1.655 cm c)W=1/2kx2=1/2(740.18)(8.90x10-2m)2=2.93 Joules

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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