Work Energy Theorem Worksheets Homework

When students walk into class, today's ranking task is already projected onto the screen at the front of the room. I choose this task because it asks students to apply their knowledge of work, specifically how work is proportional to force times distance, which is something we covered in our last class hour. The task should be a slight challenge, but this introduction is meant to get students thinking about work.

Once the students are settled, I read the instructions from the top of the activity. My reading of the instructions is to ensure students understand that class has started. I emphasize to students that they should work individually and take about 5 minutes to rank the graphs, explain their reasoning, and then assess their level of confidence. During these 5 minutes of work time, I walk around the room and informally assess how students are doing with simple glances at their work. My changes in location help students stay quiet and focused.  

When the 5 minutes are over, I reveal the answers to the students by writing them onto the front screen: C, F, E, D, B, A. I then ask if anyone got all of the answers in the correct order. This student has the right solution and is willing to share with the rest of the class. He starts by highlighting the fact that each boxcar must stop in the same distance, so work becomes proportional to only force. Then, he explains that since force equals mass times acceleration, the mass and change in velocity are what will determine the force required to stop each car. After he completes his explanation and shows a few of his calculations to support the numbers written on his paper, I end the introductory activity by asking if anyone requires further clarification. Because this activity reviews material from the previous class that we apply in today's lesson, the students keep their work to use as a reference.

Work done $w$ is given (in one-dimension) by

$$w = \int_{\mathbf x_1}^{\mathbf x_2} ~\mathbf F\cdot \mathrm d\mathbf x\tag I$$

When the force is conservative, it can be expressed by scalar potential energy function $U(\mathbf x)$ viz.

$$ \begin{align}F &= - \frac{\mathrm d}{\mathrm dx}~U\\ \implies U(\mathbf x) &= -\int_{\mathbf x_0}^{\mathbf x} \mathbf F\cdot \mathrm d\mathbf x + U(\mathbf x_0) \,, \tag{II}\end{align}$$ where $U(\mathbf x_0)$ is an arbitrary constant.

Now, as OP said

$$\Delta ~\textrm{KE} = K_2 - K_1 =\int_{\mathbf x_1}^{\mathbf x_2} ~\mathbf F\cdot \mathrm d\mathbf x $$

This can be re-written as

$$\begin{align}K_2 + \left[-\int_{\mathbf x_0}^{\mathbf x_2} ~\mathbf F\cdot \mathrm d\mathbf x\right] & = K_1 +\left[-\int_{\mathbf x_0}^{\mathbf x_1} ~\mathbf F\cdot \mathrm d\mathbf x\right]\\\implies~~ K_2 + \left[U(\mathbf x_2)-U(\mathbf x_0)\right] & = K_1 +\left[U(\mathbf x_1)-U(\mathbf x_0)\right]\\ \implies~~~~~~~~~~~~~~~~~~~~~~~~~ K_2 - K_1 & = -~U(\mathbf x_2) + U(\mathbf x_1)\\ \implies~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \Delta ~\textrm{KE} & =-~\Delta ~U\,.\tag{III} \end{align} $$

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