Question 8.9: Connected Blocks in Motion Two blocks are connected by a lig......

Conceptualize The key word rest appears twice in the problem statement. This word suggests that the configurations of the system associated with rest are good candidates for the initial and final configurations because the kinetic energy of the system is zero for these configurations.

Categorize In this situation, the system consists of the two blocks, the spring, the surface, and the Earth. The system is isolated with a nonconservative force acting. We also model the sliding block as a particle in equilibrium in the vertical direction, leading to n=m_{1} g.

Analyze We need to consider two forms of potential energy for the system, gravitational and elastic: \Delta U_{g}=U_{g f}-U_{g i} is the change in the system’s gravitational potential energy, and \Delta U_{s}=U_{s f}-U_{s i} is the change in the system’s elastic potential energy. The change in the gravitational potential energy of the system is associated with only the falling block because the vertical coordinate of the horizontally sliding block does not change. The initial and final kinetic energies of the system are zero, so \Delta K=0.

For this example, let us start from Equation 8.2

\Delta K+\Delta U+\Delta E_{\mathrm{int}}=W+Q+T_{\mathrm{MW}}{+}T_{\mathrm{MT}}+~T_{\mathrm{ET}}+~T_{\mathrm{ER}}       (8.2)

to show how this approach would work in practice. Because the system is isolated, the entire right side of Equation 8.2 is zero. Based on the physical situation described in the problem, we see that there could be changes of kinetic energy, potential energy, and internal energy in the system. Write the corresponding reduction of equation 8.2:

\Delta K+\Delta U+\Delta E_{\text {int }}=0

Incorporate into this equation that \Delta K=0 and that there are two types of potential energy:

\text { (1) } \Delta U_{g}+\Delta U_{s}+\Delta E_{\text {int }}=0

Use Equation 8.15

\Delta E_{\mathrm{int}}=f_{k}d         (8.15)

to find the change in internal energy in the system due to friction between the horizontally sliding block and the surface, noticing that as the hanging block falls a distance h, the horizontally moving block moves the same distance h to the right:

(2) \Delta E_{\mathrm{int}}=f_{k} h=\left(\mu_{k} n\right) h=\mu_{k} m_{1} g h

Evaluate the change in gravitational potential energy of the system, choosing the configuration with the hanging block at the lowest position to represent zero potential energy:

\text { (3) } \Delta U_{g}=U_{g f}-U_{g i}=0-m_{2} g h

Evaluate the change in the elastic potential energy of the system:

\text { (4) } \Delta U_{s}=U_{s f}-U_{s i}=\frac{1}{2} k h^{2}-0

Substitute Equations (2), (3), and (4) into Equation (1):

-m_{2} g h+\frac{1}{2} k h^{2}+\mu_{k} m_{1} g h=0

Solve for \mu_{k} :

\mu_{k}=\frac{m_{2} g-\frac{1}{2} k h}{m_{1} g}

Finalize This setup represents a method of measuring the coefficient of kinetic friction between an object and some surface. Notice that we do not need to remember which energy equation goes with which type of problem with this approach. You can always begin with Equation 8.2 and then tailor it to the physical situation. This process may include deleting terms, such as the kinetic energy term and all terms on the right-hand side in this example. It can also include expanding terms, such as rewriting \Delta U due to two types of potential energy in this example.