A 0.300 kg mass is attached to a 26.6 N/m spring. It is pulled
0.120 m and released. How much
Kinetic Energy does it have when it
is 0.0600 m from equilibrium?
(Unit = J)

0.144 J

Explanation:

From the question above,

Total elastic energy of the spring = Kinetic energy of the spring + potential energy of the spring

E' = E.k+Ep Equation 1

Ek = E'-Ep Equation 2

Where E' = total elastic energy, E.k = kinetic energy, E.p = potential  energy.

E'  = 1/2ke² = 1/2(26.6)(0.12²) = 0.19152 J.

Ep = 1/2ke² = 1/2(26.6)(0.06²)

Ep = 0.04788 J

Therefore,

Ek = 0.19152-0.04788

Ek = 0.144 J

i think it's 35 meters

continuous conducting path for current, power supply