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?

Answer 1

The elastic potential energy (Ep) is given by $Ep = \frac{1}{2}*k*x^2$

Data:
Ep = 5184 J
k = 16200 N/m
x (displacement) = ?

Solving:
$Ep = \frac{1}{2}*k*x^2$
$5184 = \frac{1}{2}*16200*x^2$
$5184*2 = 16200x^2$
$10368 = 16200x^2$
$16200x^2 = 10368$
$x^2 = \frac{10638}{16200}$
$x^2 = 0.64$
$x = \sqrt{0.64}$
$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\quad\checkmark$

Answer 2

Answer:the answer is 0.64 m

Explanation: