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

 Page 4 / 9

Strategy

First, we need to pick an origin for the y -axis and then determine the value of the constant that makes the potential energy zero at the height of the base. Then, we can determine the potential energies from [link] , based on the relationship between the zero potential energy height and the height at which the hiker is located.

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, $y=147\phantom{\rule{0.2em}{0ex}}\text{m}$ , so
$U\left(\text{summit}\right)=U\left(147\phantom{\rule{0.2em}{0ex}}\text{m}\right)=mgh=\left(75\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}9.8\phantom{\rule{0.2em}{0ex}}\text{N}\right)\left(147\phantom{\rule{0.2em}{0ex}}\text{m}\right)=108\phantom{\rule{0.2em}{0ex}}\text{kJ}.$
3. At sea level, $y=\left(147-195\right)\text{m}=-48\phantom{\rule{0.2em}{0ex}}\text{m}$ , so
$U\left(\text{sea-level}\right)=\left(75\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}9.8\phantom{\rule{0.2em}{0ex}}\text{N}\right)\left(-48\phantom{\rule{0.2em}{0ex}}\text{m}\right)=-35.3\phantom{\rule{0.2em}{0ex}}\text{kJ}.$

Significance

Besides illustrating the use of [link] and [link] , the values of gravitational potential energy we found are reasonable. The gravitational potential energy is higher at the summit than at the base, and lower at sea level than at the base. Gravity does work on you on your way up, too! It does negative work and not quite as much (in magnitude), as your muscles do. But it certainly does work. Similarly, your muscles do work on your way down, as negative work. The numerical values of the potential energies depend on the choice of zero of potential energy, but the physically meaningful differences of potential energy do not. [Note that since [link] is a difference, the numerical values do not depend on the origin of coordinates.]

Check Your Understanding What are the values of the gravitational potential energy of the hiker at the base, summit, and sea level, with respect to a sea-level zero of potential energy?

35.3 kJ, 143 kJ, 0

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,

$\text{Δ}U=\text{−}{W}_{AB}=\frac{1}{2}k\left({x}_{B}^{2}-{x}_{A}^{2}\right),$

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

$U\left(x\right)=\frac{1}{2}k{x}^{2}+\text{const}.$

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.)

Spring potential energy

A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm. (a) How much elastic potential energy does the spring contribute when its length is 23 cm? (b) How much more potential energy does it contribute if its length increases to 26 cm?

Strategy

When the spring is at its unstretched length, it contributes nothing to the potential energy of the system, so we can use [link] with the constant equal to zero. The value of x is the length minus the unstretched length. When the spring is expanded, the spring’s displacement or difference between its relaxed length and stretched length should be used for the x -value in calculating the potential energy of the spring.

Solution

1. The displacement of the spring is $x=23\phantom{\rule{0.2em}{0ex}}\text{cm}-20\phantom{\rule{0.2em}{0ex}}\text{cm}=3\phantom{\rule{0.2em}{0ex}}\text{cm}$ , so the contributed potential energy is $U=\frac{1}{2}k{x}^{2}=\frac{1}{2}\left(4\phantom{\rule{0.2em}{0ex}}\text{N/cm}\right){\left(3\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{2}=0.18\phantom{\rule{0.2em}{0ex}}\text{J}$ .
2. When the spring’s displacement is $x=26\phantom{\rule{0.2em}{0ex}}\text{cm}-20\phantom{\rule{0.2em}{0ex}}\text{cm}=6\phantom{\rule{0.2em}{0ex}}\text{cm}$ , the potential energy is $U=\frac{1}{2}k{x}^{2}=\frac{1}{2}\left(4\phantom{\rule{0.2em}{0ex}}\text{N/cm}\right){\left(6\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{2}=0.72\phantom{\rule{0.2em}{0ex}}\text{J}$ , which is a 0.54-J increase over the amount in part (a).

Significance

Calculating the elastic potential energy and potential energy differences from [link] involves solving for the potential energies based on the given lengths of the spring. Since U depends on ${x}^{2}$ , the potential energy for a compression (negative x ) is the same as for an extension of equal magnitude.

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