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?

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 =

K = 19.5 N/cm

e = 1.39cm

E = x 19.5 x 1.39

E = 13.55 J

The energy to stretch the spring for 6.93cm is

E = 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

i don't know that i agree with that but what's the question ?

answer: the potential energy decreases from 379 j to 243 j

explanation:

for the law of conservation of energy, the total energy in a system must be conserved. the total energy of the system is given by the sum of kinetic (k) and potential (u) energy:

e = k + u

the total energy is e=675 j, and this does not change.

initially, the kinetic energy is k=296 j, so the potential energy is

at the end, the kinetic energy is k=432 j, so the potential energy is

so, we see that the potential energy has decreased from 379 j to 243 j.