10.4.2 A Strategy for Energy Problems

To say that energy is constant or conserved is to say that the final energy, after an interaction has occurred, equals the initial energy: $(E_{sys})_{i} = (E_{sys})_{f}$ . This idea is the basis of a problem-solving strategy for energy problems:

PROBLEM-SOLVING STRATEGY 10.1
Energy-conservation problems

MODEL Define the system so that there are no external forces or so that any external forces do no work on the system. If there’s friction, bring both surfaces into the system. Model objects as particles and springs as ideal.

VISUALIZE Draw a before-and-after pictorial representation and an energy bar chart. A free-body diagram may be needed to visualize forces.

SOLVE If the system is both isolated and nondissipative, then the mechanical energy is conserved:

K_{i} + U_{i} = K_{f} + U_{f}

where $K$ is the total kinetic energy of all moving objects and $U$ is the total potential energy of all interactions within the system. If there’s friction, then

K_{i} + U_{i} = K_{f} + U_{f} + Δ E_{th}

where the thermal energy increase due to friction is $Δ E_{th} = f_{k} Δ s$ .

REVIEW Check that your result has correct units and significant figures, is reasonable, and answers the question.

Workbook Exercise 14

EXAMPLE 10.7 The speed of a pendulum

A pendulum is created by attaching one end of a 78-cm-long string to the ceiling and tying a 150 g steel ball to the other end. The ball is pulled back until the string is 60° from vertical, then released. What is the speed of the ball at its lowest point?

MODEL Let the system consist of the earth and the ball. The tension force, like a normal force, is always perpendicular to the motion and does no work, so this is an isolated system with no friction. Its mechanical energy is conserved.

VISUALIZE FIGURE 10.16 shows a before-and-after pictorial representation, where we’ve placed the zero of potential energy at the lowest point of the ball’s swing. Trigonometry is needed to determine the ball’s initial height.

The figure illustrates the before and after pictorial representations of the swinging of a pendulum.

FIGURE 10.16 Full Alternative Text

SOLVE Conservation of mechanical energy is

K_{i} + U_{Gi} = 0 + m g y_{0} = K_{f} + U_{Gf} = \frac{1}{2} m v_{1}^{2} + 0

Thus the ball’s speed at the bottom is

v_{1} = \sqrt{2 g y_{0}} = \sqrt{2 (9.80 {m/s}^{2}) (0.39 m)} = 2.8 m / s

The speed is exactly the same as if the ball had simply fallen 0.39 m.

REVIEW To solve this problem directly from Newton’s laws of motion requires advanced mathematics because of the complex way the net force changes with angle. But we can solve it in one line with an energy analysis!