Answer:
a) F = 680 N, b) W = 215 .4 J , c) F = 1278.4 N
Explanation:
a) Hooke's law is
F = k x
To find the displacement (x) let's use the elastic energy equation
[tex]K_{e}[/tex] = ½ k x²
k = 2 [tex]K_{e}[/tex] / x²
k = 2 85.0 / 0.250²
k = 2720 N / m
We replace and look for elastic force
F = 2720 0.250
F = 680 N
b) The definition of work is
W = ΔEm
W = [tex]K_{ef}[/tex] - [tex]K_{eo}[/tex]
W = ½ k ( [tex]x_{f}[/tex]² - x₀²)
The final distance
[tex]x_{f}[/tex] = 0.250 +0.220
[tex]x_{f}[/tex] = 0.4750 m
We calculate the work
W = ½ 2720 (0.47² - 0.25²)
W = 215 .4 J
We calculate the strength
F = k [tex]x_{f}[/tex]
F = 2720 0.470
F = 1278.4 N