10.3.1 Including Gravity

Now that you see how the basic energy model works, it’s easy to extend it to new situations. If a problem has both a spring and a vertical displacement, we define the system so that both the gravitational interaction and the elastic interaction are inside the system. Then we have both elastic and gravitational potential energy. That is,

U = U_{G} + U_{Sp}

(10.20)

You have to be careful with the energy accounting because there are more ways that energy can be transformed, but nothing fundamental has changed by having two potential energies rather than one.

And we know how to include the increased thermal energy if there’s friction. Thus for a system that has gravitational interactions, elastic interactions, and friction, but no external forces that do work, the energy principle is

Δ E_{sys} = Δ K + Δ U_{G} + Δ U_{Sp} + Δ E_{th} = 0

(10.21)

or, in conservation form,

K_{i} + U_{Gi} + U_{Sp i} = K_{f} + U_{Gf} + U_{Sp f} + Δ E_{th}

(10.22)

This is looking a bit more complex as we have more and more energies to keep track of, but the message of Equations 10.21 and 10.22 is both simple and profound: For a system that has no other interactions with its environment, the total energy of the system does not change. It can be transformed in many ways by the interactions, but the total does not change.

EXAMPLE 10.6 A spring-launched block

Your lab assignment for the week is to devise an innovative method to determine the spring constant of a spring. You see several small blocks of different mass lying around, so you decide to measure how high the compressed spring will launch each of the blocks. You and your lab partners realize that you need to compress the spring the same amount each time, so that only the mass is varying, and you choose to use a compression of 4.0 cm. You decide to measure height from the point on the compressed spring at which the block is released. Four launches generate the data in the table:

Mass (g)	Height (m)
50	2.07
100	1.11
150	0.65
200	0.51

What value will you report for the spring constant?

MODEL Let the system consist of the earth, the block, the spring, and the floor, so there will be two potential energies. There’s no friction and we’ll assume no drag; hence the mechanical energy of this system is conserved. Model the spring as ideal.

VISUALIZE FIGURE 10.13 shows a pictorial representation, including an energy bar chart. We’ve chosen to place the origin of the coordinate system at the point of launch. The projectile reaches height $y_{1} = h$ , at which point $v_{1} = 0 m/s$ .

The figure illustrates the before and after pictorial representation of the releasing of a block from a spring and the corresponding energy bar chart.

FIGURE 10.13 Full Alternative Text

SOLVE You might think we would need to find the block’s speed as it leaves the spring. That would be true if we were solving this problem with Newton’s laws of motion. But with an energy analysis, we can compare the system’s pre-launch energy to its energy when the block reaches its highest point, completely bypassing the launch speed. The block certainly has kinetic energy during the motion, but the net energy transfer, shown on the energy bar chart, is from elastic potential energy to gravitational potential energy.

With both elastic and gravitational potential energy included, the energy principle is

\begin{matrix} K_{i} + U_{Gi} + U_{Sp i} & = & 0 + 0 + \frac{1}{2} k {(Δ y_{0})}^{2} \\ = & K_{f} + U_{Gf} + U_{Sp f} = 0 + m g y_{1} + 0 \end{matrix}

The block travels to position $y_{1}$ , but the end of the spring does not! Be careful in spring problems not to mistake the position of an object for the position of the end of the spring; sometimes they are the same, but not always. Here the final elastic potential energy is that of an empty, unstretched spring: zero. Solving for the height, we find

y_{1} = h = \frac{k {(Δ y_{0})}^{2}}{2 m g} = \frac{k {(Δ y_{0})}^{2}}{2 g} \times \frac{1}{m}

The first expression is correct as an algebraic expression, but here we want to use the result to analyze an experiment in which we measure $h$ as $m$ is varied. By isolating the mass term, we see that plotting $h$ versus $1 / m$ (that is, using $1 / m$ as the $x$ -variable) should yield a straight line with slope $k {(Δ y_{0})}^{2} / 2 g$ .

FIGURE 10.14 is a graph of $h$ versus $1 / m$ , with masses first converted to kg. The graph is linear and the best-fit line has a $y$ -intercept very near zero, confirming our analysis of the situation. The experimentally determined slope is $0.105 m kg$ , with the units determined by rise over run. Thus the experimental value of the spring constant is

k = \frac{2 g}{{(Δ y_{0})}^{2}} \times slope = 1290 N / m

A graph of the block height (h) versus the inverse of its mass (1 over m) is shown.

FIGURE 10.14 Full Alternative Text

REVIEW 1290 N/m is a reasonably stiff spring, but that’s to be expected if you’re launching blocks a meter or more into the air.

10.3.1 Including Gravity

EXAMPLE 10.6 A spring-launched block

FIGURE 10.13 Pictorial representation of the experiment.

FIGURE 10.14 Graph of the block height versus the inverse of its mass.