Aspring has a force constant of 310.0 n/m. (a) determine the potential energy stored in the spring when the spring is stretched 4.11 cm from equilibrium. j (b) determine the potential energy stored in the spring when the spring is stretched 3.17 cm from equilibrium. j (c) determine the potential energy stored in the spring when the spring is unstretched. j

(a) 0.2618 J

(b)  0.1558 J

(c) 0 J

Explanation:

from Hook's Law,

The energy stored in a stretched spring = 1/2ke²

Ep = 1/2ke² Equation 1

Where k = spring constant, e = extension, E p = potential energy stored in the spring.

(a) When The spring is stretched to 4.11 cm,

Given: k = 310 N/m, e = 4.11 cm = 0.0411 m

Substituting these values into equation 1

Ep = 1/2(310)(0.0411)²

Ep = 155(0.0016892)

Ep =155×0.0016892

Ep = 0.2618 J.

(b) When the spring is stretched 3.17 cm

e = 3.17 cm = 0.0317 m.

Ep = 1/2(310)(0.0317)²

Ep = 155(0.0317)²

Ep = 155(0.0010049)

Ep = 0.155758 J

Ep ≈ 0.1558 J.

(c) When the spring is unstretched,

e = 0 m, k = 310 N/m

Ep = 1/2(310)(0)²

Ep = 0 J.

the body's situation won't change the forces are equal opposit and on the same body

there are no atoms