Answer:
(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.
