When a 5.00 kg object is hungvertically on a certain light spring that obeys Hoo

Question

When a 5.00 kg object is hungvertically on a certain light spring that obeys Hooke's law, thespring stretches 2.80 cm.

(a) If the 5.00 kgobject is removed, how far will the spring stretch if a 1.50 kgblock is hung on it?
1 cm

(b) How much work must an external agent do to stretch the samespring 4.00 cm from its unstretched position?
2 J (a) If the 5.00 kgobject is removed, how far will the spring stretch if a 1.50 kgblock is hung on it?
1 cm

(b) How much work must an external agent do to stretch the samespring 4.00 cm from its unstretched position?
2 J

Explanation / Answer

mass m = 5 kg streachness x = 2.8 cm = 0.028 m at equilibrium position mg = kx from this sprinf constant k = mg / x                                        = 1750 N / m (a). mass M = 1.5 kg we know kX = Mg from this streachness X = Mg / k                                    = 0.0084 m = 0.84 cm (b). Work = ( 1/ 2) k x ' ^ 2                = 0.5 * 1750 * ( 0.04 ) ^ 2    Since streachness x '= 4 cm = 0.04 m                = 1.4 J

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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