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guapka [62]
3 years ago
13

A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 72.0 J. What compression (as mea

sured from equilibrium) would result in 100 J being stored in this case?
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
6 0

Answer:14.14 cm

Explanation:

Given

Spring Compression x=12 cm

Potential energy Stored in spring=72 J

Suppose k is the spring constant of spring

Potential Energy of spring is given by =\frac{kx^2}{2}

\frac{k(0.12)^2}{2}=72

k(0.12)^2=144

k=10,000 N/m

k=10 kN/m

for 100 J energy

\frac{k(x_0)^2}{2}=100

10\times 10^3\cdot (x_0)^2=200

(x_0)^2=2\times 10^{-2}

x_0=0.1414

x_0=14.14 cm

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