Question

A vertically hanging spring is loaded with a 500.0 gram weight. The spring, which is originally 4.50 cm long, stretches to a length of 5.60 cm. a) What is the spring constant? b) What is the potential energy of the spring?

Answer 1

(a) The spring constant which will be given as :

using an obey's hooke law, we have F = k x m g = k x

k = (0.5 kg) (9.8 m/s²) / [(0.056 m) - (0.045 m)]

k = (4.9 N) / (0.011 m)

k = 445.4 N/m

(b) The potential energy of the spring which will be given as :

we know that, P.E_spring = (1/2) k x²

P.E_spring = (0.5) (445.4 N/m) (0.011 m)²

P.E_spring = 0.0269 J