# Thread: Change in Gravitational Potential Energy (Theta Involved)

1. ## Change in Gravitational Potential Energy (Theta Involved)

A 100kg man walks up a steep hill that is inclined at $30\theta$ to the horizontal. After walking $1.0km$ along this incline, his change in gravitational potential energy will be:

a) $9.8e5 J$
b) $8.5e5 J$
c) $4.9e5 J$ Correct Answer
d) $6.0e5 J$

Given
$m = 100kg$
$L_{I} = 1km = 1000m$
$30\theta$ to horizontal

Required
$E_{G}$

Analysis

I'm not exactly sure how to solve this with the given.
Do I have to use the equation $\Delta{E_{G}} = \frac{1}{2}mv^2$?

2. As I can "guess" from your question theta here stands for degree itself

Change in gravitational potential energy is = m*g*h

Where h is the distance above the initial line ,in your case its horizontal -
---------------------------------------
Distance of man above horizontal --after he has climbed 1000m along the slope of 30degree inclination = 1000 sin(30)

Put the
value of g and m in the equation