Question

when a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm. A) what is the force constant of the spring. B) if the 2.50 kg object is removed how far will the spring strecth if a 1.25 kg block is hung on it. C) how much work must an external agent do to strecth the same spring 8.00cm from its unstreched position.

Answer 1

Wo = mg = 2.50kg * 9.8N/kg = 24.5N. =
Weight of objedt.

A. Fk = Wo / d = 24.5 / 2.76 = 8.88N/cm

B. d=(1.25kg/2.50kg) * 2.76cm = 1.38cm

C. F = 8.88N/cm * 8cm = 71N.
W = Fd = 71 * 0.0888m = 6.31 Joules.

Answer 2

In your "C" answer, second line, you should have used 0.08m, not 0.0888m (that was the K constant).

Therfore, the result is W=Fd= 71*0.08= 5.68 Joules

Answer 3

2.84J