empat buah pegas masing-masing mempunyai konstanta k1=40N/m k2=60N k3 =20N/m dan k4=80n jika susunan pegas diberi beban 2 N hitunglah besar energi potensial peg

Elastisitas
- Hukum Hooke

k₁ = 40 N/m
k₂ = 60 N/m
k₃ = 20 N/m
k₄ = 80 N/m

F = 2 N

Besar Ep ....

a. Jika seri
1/kₛ = 1/k₁ + 1/k₂ + 1/k₃ + 1/k₄
1/kₛ = 1/40 + 1/60 + 1/20 + 1/80
1/kₛ = 25/240
kₛ = 240/25
kₛ = 9,6 N/m

Ep = ½ F² / kₛ
Ep = ½ × 2² / 9,6
Ep ≈ 0,208 J

b. Jika pararel
kₚ = k₁ + k₂ + k₃ + k₄
kₚ = 40 + 60 + 20 + 80
kₚ = 200 N/m

Ep = ½ F² / kₚ
Ep = ½ × 2² / 200
Ep = 0,01 J