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Arturiano [62]
3 years ago
14

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.
Physics
1 answer:
inysia [295]3 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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