Two springs with spring constants K₁=1500 N/m and K₂=3000 N/m

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Two springs with spring constants K₁=1500 N/m and K₂=3000 N/m are stretched by the same force. The ratio of potential energy stored in springs will be

Options

(a) 1:2
(b) 2:1
(c) 4:1
(d) 1:4

Correct Answer:

2:1

Explanation:

Force F=-Kxdisplacement (x),
Potential Energy = U =(1/2) Kx²
U = (1/2) K(F/K)² = (1/2)(F²/K)
U₁/U₂=(1/2)(F²/K₁)x(2K₂/F²)
K₂/K₁=3000/1500=2/1.

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Topics: Laws of Motion (103)
Subject: Physics (2479)

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