10.8.1 Energy Bar Charts Expanded

Energy bar charts can now be expanded to include the thermal energy and the work done by external forces. The energy principle, Equation 10.30, can be rewritten as

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

(10.31)

The initial mechanical energy $(K_{i} + U_{i})$ plus any energy added to or removed from the system $(W_{ext})$ becomes, without loss, the final mechanical energy $(K_{f} + U_{f})$ plus any increase in the system’s thermal energy $(Δ E_{th})$ . Remember that we have no way to determine $E_{th i}$ or $E_{th f}$ , only the change in thermal energy. $Δ E_{th}$ is always positive when the system contains dissipative forces.

EXAMPLE 10.10 Hauling up supplies

A mountain climber uses a rope to drag a bag of supplies up a slope at constant speed. Show the energy transfers and transformations on an energy bar chart.

MODEL Let the system consist of the earth, the bag of supplies, and the slope.

SOLVE The tension in the rope is an external force that does work on the bag of supplies. This is an energy transfer into the system. The bag has kinetic energy, but it moves at a steady speed and so $K$ is not changing. Instead, the energy transfer into the system increases both gravitational potential energy (the bag is gaining height) and thermal energy (the bag and the slope are getting warmer due to friction). The overall process is $W_{ext} \to U + E_{th}$ . This is shown in FIGURE 10.30.

The figure illustrates an energy bar chart.

FIGURE 10.30 Full Alternative Text

CHAPTER 10 CHALLENGE EXAMPLE A spring workout

An exercise machine at the gym has a 5.0 kg weight attached to one end of a horizontal spring with spring constant $80 N/m$ . The other end of the spring is anchored to a wall. When a woman working out on the machine pushes her arms forward, a cable stretches the spring by dragging the weight along a track with a coefficient of kinetic friction of 0.30. What is the woman’s power output at the moment when the weight has moved $50 cm$ if the cable tension is a constant 100 N?

MODEL This is a complex situation, but one that we can analyze. First, identify the weight, the spring, the wall, and the track as the system. We need to have the track inside the system because friction increases the temperature of both the weight and the track. The tension in the cable is an external force. The work $W_{ext}$ done by the cable’s tension transfers energy into the system, causing K, $U_{Sp}$ , and $E_{th}$ all to increase.

VISUALIZE FIGURE 10.31a is a before-and-after pictorial representation. The energy transfers and transformations are shown in the energy bar chart of Figure 10.31b.

The figure labeled (“a”) illustrates the before and after pictorial representations of the displacement of a block when a horizontal spring is stretched.

The figure labeled (b) illustrates an energy bar chart.

Full Alternative Text

SOLVE You learned in 《 SECTION 9.6 that power is the rate at which work is done and that the power delivered by force $\vec{F}$ to an object moving with velocity $\vec{v}$ is $P = \vec{F} \cdot \vec{v}$ . Here the tension $\vec{T}$ pulls parallel to the weight’s velocity, so the power being supplied when the weight has velocity $\vec{v}$ is $P = T v$ . We know the cable’s tension, so we need to use energy considerations to find the weight’s speed $v_{1}$ after the spring has been stretched to $Δ x_{1} = 50 cm$ .

The energy principle $K_{i} + U_{i} + W_{ext} = K_{f} + U_{f} + Δ E_{th}$ is

\frac{1}{2} m v_{0}^{2} + \frac{1}{2} k {(Δ x_{0})}^{2} + W_{ext} = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} k {(Δ x_{1})}^{2} + Δ E_{th}

The initial displacement is $Δ x_{0} = 0 m$ and we know that $v_{0} = 0 m / s$ , so the energy principle simplifies to

\frac{1}{2} m v_{1}^{2} = W_{ext} - \frac{1}{2} k {(Δ x_{1})}^{2} - Δ E_{th}

The external work done by the cable’s tension is

W_{ext} = T Δ r = (100 N)(0.50 m) = 50.0 J

From Chapter 9, the increase in thermal energy due to friction is

\begin{matrix} Δ E_{th} & = & f_{k} Δ r = μ_{k} m g Δ r \\ = & (0.30) (5.0 kg)(9.80 {m/s}^{2}) (0.50 m) = 7.4 J \end{matrix}

Solving for the speed $v_{1}$ , when the spring’s displacement is $Δ x_{1} = 50 cm = 0.50 m,$ we have

v_{1} = \sqrt{\frac{2 (W_{ext} - \frac{1}{2} k {(Δ x_{1})}^{2} - Δ E_{th})}{m}} = 3.6 m / s

The power being supplied at this instant to keep stretching the spring is

P = T v_{1} = (100 N)(3.6 m/s) = 360 W

REVIEW The work done by the cable’s tension is energy transferred to the system. Part of the energy increases the speed of the weight, part increases the elastic potential energy stored, and part is transformed into increased thermal energy, thus increasing the temperature. We had to bring all these energy ideas together to solve this problem.

10.8.1 Energy Bar Charts Expanded

EXAMPLE 10.10 Hauling up supplies

FIGURE 10.30 The energy bar chart for Example 10.10.

CHAPTER 10 CHALLENGE EXAMPLE A spring workout

FIGURE 10.31 Pictorial representation and energy bar chart for Chapter 10 Challenge Example.