8.1 Potential energy of a system  (Page 4/9)

Strategy

Solution

1. Let’s choose the origin for the y -axis at base height, where we also want the zero of potential energy to be. This choice makes the constant equal to zero and
   
   $U\left(\text{base}\right)=U\left(0\right)=0.$
2. At the summit, $y=147\text{m}$ , so
   
   $U\left(\text{summit}\right)=U\left(147\text{m}\right)=mgh=\left(75\times 9.8\text{N}\right)\left(147\text{m}\right)=108\text{kJ}.$
3. At sea level, $y=\left(147-195\right)\text{m}=\mathrm{-48}\text{m}$ , so
   
   $U\left(\text{sea-level}\right)=\left(75\times 9.8\text{N}\right)\left(\mathrm{-48}\text{m}\right)=\mathrm{-35.3}\text{kJ}.$

Significance

Elastic potential energy

In Work , we saw that the work done by a perfectly elastic spring, in one dimension, depends only on the spring constant and the squares of the displacements from the unstretched position, as given in [link] . This work involves only the properties of a Hooke’s law interaction and not the properties of real springs and whatever objects are attached to them. Therefore, we can define the difference of elastic potential energy for a spring force as the negative of the work done by the spring force in this equation, before we consider systems that embody this type of force. Thus,

where the object travels from point A to point B . The potential energy function corresponding to this difference is

If the spring force is the only force acting, it is simplest to take the zero of potential energy at $x=0$ , when the spring is at its unstretched length. Then, the constant is [link] is zero. (Other choices may be more convenient if other forces are acting.)