imported>Patrick at 17:58, 13 August 2011

2011-08-13T17:58:41Z

New page

The block slides down the distance 3 m and then continues to slide down some more. Let's assume that it slides some further distance <math>x</math>. It means that the total distance the block moves down the incline is <math>3 + x</math>. It moves down the height <math>h = (3 + x) \sin(30^{\circ})</math>. The spring which was initially uncompressed is now compressed by <math>x</math>. 
[[image:Helena_Pot_Energy_Ex_4_Soln.png|TOP]] 

The question is to find the distance <math>x</math> such that the total change in potential energy of the system (spring-block-Earth) is zero. To find this distance we will first compute the change of height from the triangle in the diagram above:

<math>\sin(30^{\circ}) = {h \over 3 + x}</math>

<math>h = (3 + x) \sin(30^{\circ})</math>
 
And then the change in gravitational potential energy using the equation:

<math>\Delta U_G = -m g h = -(0.1)(10)((3 + x) \sin(30^{\circ}))</math>
 
And then compute the change in the spring's potential energy:

<math>\Delta U_{sp} = \frac{1}{2} k {x_f}^2 - \frac{1}{2} k {x_i}^2 = \frac{1}{2} k x^2</math>
 
The total change in potential energy is then

<math>\Delta U_G + \Delta U_{sp} = 0</math>

<math>[-(0.1)(10)((3 + x) \sin(30^{\circ}))] + [\frac{1}{2} (4) x^2] = 0</math>

<math>-(3 + x) + 4 x^2 = 0</math>
 
Solving the quadratic equation for <math>x</math>, we find <math>x = 1</math> m. Note that we select the positive solution since the distance is always positive.

Potential Energy EX 4 - Revision history

imported>Patrick at 17:58, 13 August 2011