Question

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

Answer 1

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! =)

Answer 2

Simply use the equation E = 0.5kx^2. Here E is the elastic potential energy stored in the spring, k is the spring constant and x is the extension. We are solving for extension:

E = 0.5kx^2
x = sqrt(2E/k) = sqrt((2*5184)/16200) = 0.80m

I hope this helps you :)