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11.1.6.2: Explorations

  • Page ID
    34092
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    Exploration 1: An Operational Definition of Work

    This Exploration allows you to discover how work causes changes in kinetic energy. Restart.

    Drag "handy" to the front and/or the back of the cart to impart a force (position is given in meters and time is given in seconds). Look at the force \(\cos(\theta )\) vs. position graph as well as the final velocity.

    1. What can you say about the relationship between the force applied and the work done?
    2. How does the application of the force change the kinetic energy?
    3. What happens when the mass changes?

    Make sure to push from both sides when the cart is both stationary and moving to the left and to the right.

    Exploration 2: Two Two-Block Push

    Two blocks are pushed by identical forces (position is given in meters and time is given in seconds), each block starting at rest at the first vertical rectangle (start). The top block is twice the mass of the bottom block, \(m_{1} = 2m_{2}\). Restart. The graphs and tables are initially blank. Animation without Graphs and Tables.

    1. Which object has the greater kinetic energy when it reaches the second vertical rectangle (finish)? Why?
    2. Once you have answered the above question, click Animation with Graphs and Tables to see if you were correct. Consider both the graphs and the tables.
    3. If you were incorrect in your answer, can you figure out why you answered incorrectly? What is the correct rationale you should have used to answer this question? Use the graphs and tables where appropriate.

    Exploration 3: The Gravitational Force and Work

    A \(1\text{-kg}\) ball is subjected to the force of gravity as shown in the animation (position is given in meters and time is given in seconds). The ball starts at \(x = 0\text{ m}\) and \(y = 0\text{ m}\). You can vary the ball's initial velocity and view how this affects the motion of the ball and the ball's kinetic energy. Also shown are the graphs of force \(\cos(\theta_{y})\) vs. \(\text{position}_{y}\) and kinetic energy vs. \(\text{position}_{y}\) where \(\theta_{y}\) is the angle between the gravitational force and the ball's infinitesimal \(y\) displacement. Restart.

    With \(v_{0x} = 0\text{ m/s}\) and \(v_{0y} = 0\text{ m/s}\),

    1. In what direction does the net force on the ball point?
    2. How would you describe the kinetic energy vs. position graph?

    With \(v_{0x} = 10\text{ m/s}\) and \(v_{0y} = 0\text{ m/s}\),

    1. What is the ball's minimum kinetic energy?

    With \(v_{0x} = 10\text{ m/s}\) and \(v_{0y} = 10\text{ m/s}\),

    1. How would you describe the KE vs. position graph? Be explicit about what happens to the kinetic energy.
    2. What is the condition for the work done by gravity to be zero?
    3. What is the ball's minimum kinetic energy?

    With \(v_{0x} = -10\text{ m/s}\) and \(v_{0y} = 10\text{ m/s}\),

    1. How would you describe the kinetic energy vs. position graph? Be explicit about what happens to the kinetic energy.
    2. What is the condition for the work done by gravity to be zero?
    3. What is the ball's minimum kinetic energy?

    Exploration 4: Change the Direction of the Force Applied

    A \(20\text{-kg}\) ball has a hole with a rod passing through. The rod exerts a force as needed that constrains the ball to move along the rod. An applied force is now added (the "pulling" force) so the ball is pulled as shown (position is given in meters and time is given in seconds). The pulling force vector is shown as a red arrow, and the force makes an angle \(\theta\) with the horizontal. The velocity is given in meters/second. You may adjust the angle and/or the magnitude of the pulling force (\(F < 7\text{ N}\)). Restart.

    1. How does the work done by the pulling force change as you vary the pulling force for a constant angle?
    2. How does the work done by the pulling force change as you vary the angle for a constant pulling force?
    3. Combine your answers above to obtain a general mathematical formula for the work done on the ball due to an arbitrary pulling force.
    4. Determine the general mathematical formula for the work done by the force the rod exerts on the ball when an arbitrary force is applied to the ball.

    Exploration 5: Circular Motion and Work

    A \(1\text{-kg}\) black ball is constrained to move in a circle as shown in the animation (position is given in meters, time is given in seconds, and energy on the bar graph is given in joules). In the animation the wire is vertical and the ball is subjected to gravity (downward as usual), as well as the force of the wire. You may set the initial velocity and then play the animation. The blue arrow represents the net force acting on the mass, while the bar graph displays its kinetic energy in joules. Restart.

    1. Set the speed fast enough to get the ball over the top. Then restart and examine forces at positions near, say, \(-45^{\circ}\) and \(45^{\circ}\) (hanging straight down is \(-90^{\circ}\)). Label your forces as \(F_{g}\) (gravity), \(F_{\text{wire}}\), and \(F_{\text{net}}\).
    2. Given your force diagrams, there are positions where the speed of the ball is changing more rapidly than others. Take each of the positions you considered and rank them from highest tangential acceleration to lowest.
    3. Assume that the ball can get to \(y = 10\text{ m}\). How much kinetic energy does the ball lose in going from \(y = -10\text{ m}\) to \(y = 10\text{ m}\)? Is this independent of \(v_{0x}\) initial?
    4. What is the work done by gravity when the ball goes from \(y = -10\text{ m}\) to \(y = 10\text{ m}\)?
    5. Determine the minimum speed that the ball must have to go over the top. Once you have an answer, check it using the animation.

    Exploration 6: Forces, Path Integrals, and Work

    Move your cursor into the animation, then click-drag the crosshair cursor with the mouse. The bar graph on the right displays the work done by the force along the path. For your reference, there are circles every \(10\text{ m}\) that form a coordinate grid (position is given in meters and the result of the integral is given on the bar graph in joules). Use the "reset integral" button to re-zero the work calculation between paths. Restart.

    For each force, answer the following questions:

    1. Starting at the origin (the center, \(x = 0\text{ m}\) and \(y = 0\text{ m}\)) and moving to \(x = 0\text{ m}\) and \(y = 10\text{ m}\), what is the work done by the force?
    2. Starting at \(x = 0\text{ m}\) and \(y = 10\text{ m}\) and moving to \(x = 0\text{ m}\) and \(y = 0\text{ m}\), what is the work done by the force?
    3. Starting at the origin (the center, \(x = 0\text{ m}\) and \(y = 0\text{ m}\)) and moving to \(x = 0\text{ m}\) and \(y = -10\text{ m}\), what is the work done by the force?
    4. Starting at the origin (the center, \(x = 0\text{ m}\) and \(y = 0\text{ m}\)) and moving to \(x = 10\text{ m}\) and \(y = 0\text{ m}\), what is the work done by the force?
    5. Starting at the origin (the center, \(x = 0\text{ m}\) and \(y = 0\text{ m}\)) and moving to \(x = -10\text{ m}\) and \(y = 0\text{ m}\), what is the work done by the force?
    6. Starting at the origin (the center, \(x = 0\text{ m}\) and \(y = 0\text{ m}\)), choosing your own path around the animation and ending back at the origin, what is the work done by the force?

    When you are through, feel free to experiment with forces of your own choosing.

    Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.


    This page titled 11.1.6.2: Explorations is shared under a CC BY-NC-ND license and was authored, remixed, and/or curated by Wolfgang Christian, Mario Belloni, Anne Cox, Melissa H. Dancy, and Aaron Titus, & Thomas M. Colbert.