Understanding Spring Compression: How Doubling It Affects Stored Energy

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Doubling the compression of a spring increases the stored energy by a factor of four, not two. The potential energy of a spring is calculated using the formula Fs = 1/2 kx^2. When the compression (x) is doubled, the equation becomes Fs2 = 1/2 k(2x)^2, which simplifies to four times the original energy. This highlights the importance of understanding the relationship between compression and energy storage in springs. The discussion clarifies that the spring constant (k) remains unchanged while the compression is what affects the energy increase.
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If compression of a spring is doubled, the stored energy increased by a factor of?

I think that the stored energy is increased by a factor of 2.

Right?
 
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What is the equation for a spring's potential energy?
 
Fs1=1/2 kx^2 so if k = 25 v=4
= 200

Fs2= k=50 v=4
=400

so by factor of 2
 
Ah, you are doubling the wrong thing. k is the spring constant. The question is asking about changing x by a factor of 2.

F_{s1} = .5kx^2

F_{s2} = .5k(2x)^2

\frac{F_{s2}}{F_{s1}} = ?
 
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