Question

When a 2.10-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.52 cm.(a) What is the force constant of the spring?N/m(b) If the 2.10-kg object is removed, how far will the spring stretch if a 1.05-kg block is hung on it?cm(c) How much work must an external agent do to stretch the same spring 6.40 cm from its unstretched position?3 J

Answer 1

(a) The force constant of the spring is 817.5 N/m.

(b) The spring will stretch for 1.26 cm if a 1.05 kg block is hung on it.

(c) The work done by the external agent to stretch the spring by 6.4 cm is

1.67 J.

(a) Given that,

Mass of the spring = 2.1 kg

The spring stretches = 2.52 cm = 0.0252 m

Force constant k = Force applied / extension produced

k = (2.10kg * 9.81 N/kg) / 0.0252 m

k = 817.5 N/m

(b) If the 2.10 kg object is removed and a 1.05 kg block is hung on the spring.

Extension = F/k = (1.05 kg * 9.81) / 817.5 =  0.0126 m = 1.26 cm

(c) The work done by the external agent to stretch the spring by 6.4 cm is,

W = average force used * distance

W = 1/2 * k* e * e = 1/2 k*e²

W = 1/2 * 817.5 * (0.064)² = 1.67 J

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