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2 years ago

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A spring of spring constant k=8.25N/m is displaced from equilibrium by a distance of 0.150m. What is the stored energy in the fo

rm of spring potential energy? Show your work.

1 answer:

inysia [295]2 years ago

4 0

Answer:

0.0928J

Explanation:

the pulling force of spring F=-kx

where x is the displacement from equilibrium position.

energy stored:

$\int\limits^x_0 {-F} \, dx \\=\int\limits^x_0 {kx} \, dx \\\\=\frac{kx^{2} }{2}$

*** Its fine if you know nothing about calculus. Just apply the equation

$U=\frac{kx^{2} }{2}$

where U is the potential energy of the spring***

put x=0.150, $U=\frac{8.25}{2}$ × $0.150^{2}$ = 0.0928J (corr. to 3 sig. fig.)

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