# Question 0affe

Dec 4, 2017

Only the Potential Energy Difference between two points has physical meaning.

The potential energy at a point has no physical meaning and one can arbitrarily pick a point as reference and assign zero potential energy for that point.

#### Explanation:

The most important point to remember while dealing with potential energy is - it is the difference in potential energy between two points that is measurable and not the potential energy itself.

In fact no two observers will agree on the value of potential energy at a point but all will agree on the difference in potential energy between two points in space. I may assign the ground I am standing on as zero potential, but for someone one floor above me it is a below and that person might assign a negative potential and someone below me sees it as above and assign a positive potential.

This Problem:
h_i = 0; \qquad h_f=-h;\qquad v_i=0m.s^{-1};\qquad v_f=?

Mechanical Energy Conservation: \qquad \DeltaE=\DeltaU + \DeltaK = 0;

K_i = 0; \qquad K_f = 1/2mv_f^2;\qquad U_i = 0; \qquad U_f=-mgh;

\DeltaK = - \DeltaU;\qquad (K_f-K_i) = - (U_f-U_i);

\DeltaK = K_f - K_i = 1/2mv_f^2 - 0 = 1/2mv_f^2;

\DeltaU = U_f - U_i = mgh_f - 0 = (-mgh) - 0 = -mgh;

\DeltaK = -\DeltaU; \qquad \rightarrow 1/2mv_f^2 = -(-mgh);

${v}_{f} = \setminus \sqrt{2 g h}$

From a Different Perspective: From the perspective of the bottom of the ramp.

h_i = h; \qquad h_f=0;\qquad v_i=0m.s^{-1};\qquad v_f=?

Mechanical Energy Conservation: \qquad \DeltaE=\DeltaU + \DeltaK = 0;

K_i = 0; \qquad K_f = 1/2mv_f^2;\qquad U_i = +mgh; \qquad U_f=0;

\DeltaK = - \DeltaU;

\DeltaK = K_f - K_i = 1/2mv_f^2 - 0 = 1/2mv_f^2;

\DeltaU = U_f - U_i = mgh_f - mgh_i = 0 - (+mgh) = -mgh;

\DeltaK = -\DeltaU; \qquad \rightarrow 1/2mv_f^2 = -(-mgh);#

${v}_{f} = \setminus \sqrt{2 g h}$