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/s^{2}) / [(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 x^{2}

P.E_{spring} = (0.5) (445.4 N/m) (0.011
m)^{2}

**P.E _{spring} = 0.0269 J**

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