Question

When a 2.60-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.86 cm. (a) What is the force constant of the spring? N/m (b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it? cm (c) How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?

Answer 1

Answer:

a)  k = 891.82 N/m

b) e = 0.0143 m = 1.43 cm

c) W = 5.02 J

Explanation:

Step 1: Data given

Mass = 2.60 kg

the spring stretches 2.86 cm = 0.0286

Step 2: What is the force constant of the spring?

Force constant, k = force applied / extension produced  

k = (2.60kg * 9.81N/kg) / 0.0326 m

k = 891.82 N/m

b)  If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it

Extension = F/k = (1.30 kg * 9.81) / 891.82 =  0.0143 m = 1.43 cm

Half the mass means half the extension

c) How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?

W = average force used * distance

W = 1/2 * k*e * e = 1/2 k*e²  

W = 1/2 * 891.82 * (0.075)² = W = 5.02 J