Hello!
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?
0.57 m
0.64 m
0.80 m
1.25 m
Data:
\(E_{pe}\:(elastic\:potential\:energy) = 5184\:J\)
\(K\:(constant) = 16200\:N/m\)
\(x\:(displacement) =\:?\)
For a spring (or an elastic), the elastic potential energy is calculated by the following expression:
\(E_{pe} = \dfrac{k*x^2}{2}\)
Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.
Solving:
\(E_{pe} = \dfrac{k*x^2}{2}\)
\(5184 = \dfrac{16200*x^2}{2}\)
\(5184*2 = 16200*x^2\)
\(10368 = 16200\:x^2\)
\(16200\:x^2 = 10368\)
\(x^{2} = \dfrac{10368}{16200}\)
\(x^{2} = 0.64\)
\(x = \sqrt{0.64}\)
\(\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark\)
Answer:
The displacement of the spring = 0.8 m (or 0.80 m)
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I Hope this helps, greetings ... Dexteright02! =)
