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Makovka662 [10]
3 years ago
8

You compress a spring by a distance of 0.2 m. The spring has a spring constant of 37 N/m. When you release the spring, it snaps

back. What is the kinetic energy of the spring as it reaches its natural length?
Physics
2 answers:
Serggg [28]3 years ago
7 0
At that point it is no longer trying to uncompress nor is it trying to stretch.  This is the same thing as a pendulum at the bottom of its swing, no longer falling but not yet rising against gravity.  Thus the kinetic energy there is the same as the potential energy when it is compressed.  The energy of compression is
\frac{1}{2}k x^{2}
This gives E=0.5(37)(0.2)²=0.74J
This is the same as the kinetic energy when it is at natural length
Angelina_Jolie [31]3 years ago
7 0

Answer:

0.74 J

Explanation:

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Explanation:

given that data in the question; as its interpreted in the diagram below;

from the cosine rule, we know that;

a² = b² + c² - 2bc  

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169 = 36 + 324 - 216cos∅

169 = 360 - 216cos∅

216cos∅ = 360 - 169

216cos∅ = 191

cos∅ = 0.8842

∅ = cos⁻¹ ( 0.8842 )

∅ = 27.8° ≈ 28°  {nearest whole number}

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