imported>Patrick at 15:24, 13 August 2011

2011-08-13T15:24:40Z

New page

When released, the block of mass moves down and therefore the gravitational potential energy of the system (Earth-block-spring) decreases. Consequently, the change in gravitational potential energy is negative. While the block slides down some distance <math>x</math>, the spring stretches the same distance. Consequently, the potential energy of the spring increases. The change of the gravitational potential energy can be computed using:

<math>\Delta U = -m g h = -(3)(10) x = -30 x</math>
<br><br>
The initial extension of the spring is <math>x_i = 0</math> and the final extension of the spring is <math>x_f = x</math>. We can compute the change in potential energy of the spring as follows:

<math>\Delta U = \frac{1}{2} k {x_f}^2 - \frac{1}{2} k {x_i}^2 = \frac{1}{2} k x^2 = 10 x^2</math>
<br><br>
The total potential energy in this case is equal to zero and so

<math>\Delta U = -30 x + 10 x^2 = 0</math>

Solving for <math>x</math>, we find <math>x = 3</math> m.

Potential Energy EX 3 - Revision history

imported>Patrick at 15:24, 13 August 2011