Question

A spring with k = 19.5 N/cm is initially stretched 1.39 cm from its equilibrium length. a) How much more energy is needed to further stretch the spring to 6.93 cm beyond its equilibrium length?

Answer 1

Answer:

Energy needed = 54.02 J

Explanation:

the Energy in an elastic spring from hookes law is given  as

F= ke , therefore the energy (E) is

E = $\frac{1}{2} Ke^{2}$

K = 19.5 N/cm

e = 1.39cm

E = $\frac{1}{2}$ x 19.5 x 1.39

E = 13.55 J

The energy to stretch the spring for 6.93cm is

E = $\frac{1}{2}$ x 19.5 x 6.93

E = 67.57 J

The more energy needed for the further stretch is

67.57 - 13.55

Energy needed = 54.02 J