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Apr 3, 2024

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Module 2: Week 6 Work and Kinetic Energy

Module 2, Week 6 1 Challenge Problem: A Ball Suspended in an Accelerating Car This activity should take about 5 minutes to complete. For this problem consider the ball hanging from a string attached to the roof of a car, as shown in the diagram: What is the angle or the ball on the string with respect to vertical if the car accelerates with ࠵౎࠵? = 2 m/s 2 toward the left? To correctly complete this problem you must develop a set of sequential plan steps that includes: 1. Draw a correct free-body diagram. 2. Execute the “Big-Idea-Justification-Plan” technique correctly. 3. Find the correct angle.

Module 2, Week 6 2 Work: Pulling a Box with a Rope Activity 1: This activity should take about 10 minutes to complete. You need to move a large crate from one end of the block to the other. You plan to pull the crate using a rope along the rough sidewalk. The crate has a mass ࠵ీ࠵? = 50 kg and you pull on the rope at an angle ࠵༃࠵? = 25 ∘ with respect to the horizontal. A friend of yours determined that the coefficient of kinetic friction between the crate and the sidewalk is ࠵༇࠵? ࠵ౘ࠵? = 0.15 . You pull the crate with constant speed. Categorize the following questions according to the Big Ideas you would use to solve them most efficiently. Provide your justification in complete sentences. Question Question 1 : What is the tension in the rope? Big Idea & Justification: Question 2 : What is the normal force between the crate and the sidewalk? Big Idea & Justification: Question 3 : What is the force of friction between the crate and the sidewalk? Big Idea & Justification: Question 4 : How much work is done by the force of friction between the crate and the surface in 3 m travel? Big Idea & Justification: Question 5 : How much work do you do moving the crate 3 m? Big Idea & Justification: forces in ID-box pulled at o by rope N21 -box pulled by rope at o N3L - normal force equal & opposite net force experienced by sidewalk from crate N3L - Of is result of sidewalk on crate Work-forces acting over a distance Work - 7 is in direction of motion so crate does work

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Module 2, Week 6 3 Work as pairs to develop a complete set of sequential plan steps for answering Questions 4 or 5. One pair should work on Question 4 and the other pair should work on Question 5. Trade your plan with the other pair in your discussion team. Use your teammates plan to answer Question 4 or 5. Make sure you execute each plan step as written. Were you able to answer the question with your teammates’ plan? What changes would you make?

Module 2, Week 6 4 Exploring Forces, Friction, and Work Activity 2: This activity should take about 15 minutes to complete. Launch the following simulation and launch the Acceleration simulation: https://phet.colorado.edu/sims/html/forces-and-motion-basics/latest/forces-and-motion- basics_en.html Set up the simulation so that all values are shown:

Module 2, Week 6 5 Part 1: The Frictionless Surface Set up the simulation for a frictionless surface. Question 1 : For how many seconds must you apply a 200 N force to accelerate the 50 kg crate from ࠵ౣ࠵? 0 = 0 m/s to ࠵ౣ࠵? = 20 m/s? Apply 200 N of force to the 50 kg crate in the simulation for the number of seconds you calculated and then stop the force on the crate. 1. What is the speed of the crate? 2. How does the speed change as time passes? Question 2: Now apply the 200 kg refrigerator to the top of the crate. What happens to the speed? What Big Ideas and Justifications explain how the speed changes with the application of the refrigerator? Once you have answered these questions, reset the simulation to the original settings and make sure all values are shown. 200 = 500 V = Vo + a + t = S 4 = a 20 = 4 + decreases Work-KE theore - 7 does work and changes speed

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Module 2, Week 6 6 Part 2: The Maximum Friction Surface Setup the simulation for “lots” of friction. Question 1: What is the coefficient of static friction between the crate and the surface? Work together to develop a plan to experimentally (through the simulation) find the coefficient of static friction between the crate and the surface. Execute your plan and find the coefficient of static friction. Question 2: What happens once static friction has been overcome? What is the coefficient of kinetic friction between the crate and the surface? Question 3: How much work is done by the force of kinetic friction as the crate moves 25 meters? Question 4: How much work is done by the person pushing as the crate moves 25 meters? Now add the 80 kg person to the top of the crate. Question 5: How much work is done by the force of friction as the crate slows to a stop?

Module 2, Week 6 7 Kinetic Energy: The Energy of Motion Activity 3: This activity should take about 15 minutes to complete. A car is initially traveling down Green Street with a speed ࠵ౣ࠵? ࠵ౖ࠵? = 2 m/s. The driver needs to change speeds. The driver presses the accelerator and the engine does 1000 J of work to accelerate the car. During the acceleration, the car travels Δ࠵౥࠵? = 3 m. Work together as a discussion team to categorize the following questions according to the Big Ideas you would use to solve them most efficiently. Question 1 : If the car weighs 6600 N what is the change in kinetic energy of the car as it accelerates to the new speed, ࠵ౣ࠵? ࠵౛࠵?࠵౛࠵?࠵౛࠵? ? Question 2 : What is the new speed, ࠵ౣ࠵? ࠵౛࠵?࠵౛࠵?࠵౛࠵? ? Question 3 : How long did it take to change the car’s speed?

Module 2, Week 6 8 Work together as a discussion team to answer each question. Record your results in the table: Question Question 1 : If the car weighs 6600 N what is the change in kinetic energy of the car as it accelerates to the new speed, ࠵ౣ࠵? ࠵౛࠵?࠵౛࠵?࠵౛࠵? ? Plan Steps: Result: Question 2 : What is the new speed, ࠵ౣ࠵? ࠵౛࠵?࠵౛࠵?࠵౛࠵? ? Plan Steps: Result: Question 3 : How long did it take to change the car’s speed? Plan Steps: Result

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Module 2, Week 6 9 Using the Work-Kinetic Energy Theorem Activity 4 This activity should take about 20 minutes to complete. You are watching a box slide along a flat, level surface. You decide it will be fun to figure out the coefficient of kinetic friction between the box and the surface. As the box passes the point where you are sitting you measure its speed as ࠵ౣ࠵? ࠵ౖ࠵? = 2 m/s . Five seconds later, you realize that the box has come to rest. You pick up the box and weigh it. You find it has mass ࠵ౚ࠵? = 10 kg . Answer the following question using two different techniques: Question : What is the coefficient of kinetic friction, ࠵༇࠵? ࠵ౘ࠵? , between the surface and the box? Technique 1: Kinematics & Newton’s Laws Use forces, Newton’s Laws, and Kinematics to answer the question. Work together to identify the Big Ideas and Justifications that apply. Big Ideas: Justification: Now work together to develop a sequential plan to find the coefficient of kinetic friction, ࠵༇࠵? ࠵ౘ࠵? . Plan: Vf-o

Module 2, Week 6 10 Technique 2: Work-Kinetic Energy Theorem & Kinematics Use the Work–Kinetic Energy Theorem to answer the question. Work together to identify the Big Ideas and Justifications that apply. Big Ideas: Justification: Now work together to develop a sequential plan to find the coefficient of kinetic friction, ࠵༇࠵? ࠵ౘ࠵? . Plan: Compare the two techniques. Which technique do you prefer and why?

13 Force: Normal force Justification: The object is pushed by a surface to prevent is from sinking in. The object tends to sink in due to its weight. The normal force and the weight is the action-reaction force pair. There is no fixed value of the normal force; it changes depending on the external force(s) and the state of motion. Back to Table 1 . 14 Force: Tension Justification: The object is pulled by a rope (or a string). A tension force points along the direction of a rope under stress when being pulled by a force or by a weight. The tension forces at the two ends of the rope have the same magnitude, but point in opposite directions. Note that the tension force does not have a fixed value; it changes depending on the external force(s) and the state of motion. Back to Table 1 . 15 Force: Friction Justification: The object is on a surface with friction. Friction is the resistance of an object to slide against another object. There are two types of friction, depending on the state of motion. When an object slides on a rough surface, it experiences a kinetic friction: F k = μ k N where N is the normal force and mu k is the coe ffi cient of kinetic friction. The direction of frictional force is opposite to the direction of local motion. Friction acts to slow down the motion. When an object is not moving on a surface, but is pushed or pulled by an external force which tried to start the motion, then the static friction acts to cancel the external force, up to a maximum force: F s  μ s N where N is the normal force and mu s is the coe ffi cient of static friction. Note that μ s and μ k depends on the type of surface, and μ k < μ s for same object on same surface. Back to Table 1 . 16 Force: Spring Justification: The object is attached to a spring and causing the spring to deform by stretching or compressing the spring. The object experiences a force from the deformed spring. Equations: F k = - kx where k is the spring constant, and x is the displacement from the point of equilibrium. The negative sign indicates the spring force points opposite to the direction of displacement. The spring force changes; it scales linearly with the amount of displacement. This is called the Hook’s law. Back to Table 1 . 17 Work Justification: The force in the question is applied for a specified distance. Equations: The work, W , done by the applied force, ~ F , over a distance of ~ d is W = ~ F · ~ d = Fd cos ✓ where ✓ is the angle between the force vector, ~ F , and the displacement vector, ~ d . The work is a scalar quantity. Back to Table 2 . 5

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18 Kinetic energy Justification: The energy of motion is needed to solve this problem. Equations: The kinetic energy, E k , of a moving object is E k = 1 2 mv 2 where m is the mass and the v is the velocity of the moving object. The kinetic energy is a scalar quantity. Back to Table 2 . 19 Work-Kinetic energy theorem Justification A force does work and causes a change in the speed of an object. Equations: Work changes the kinetic energy: W = Δ E k = E f k - E i k where E f k and E f i are the final and initial kinetic energy of the object. Back to Table 2 . 20 Potential energy Justification: Stored energy is important in this problem. Equations: An object stores a potential energy, U , under the earth’s gravitational field: U = mgh where m is the mass of the object, g the gravitational acceleration of 9.8 m/s 2 , h is the height of the object. Back to Table 2 . 21 Conservation of energy Justification: Energy changes forms in this problem, but there are no external forces doing work. Equations: If there are not non-conservative forces doing work, then the total mechanical energy of the system, E tot = E i k + U i = E f k + U f , remains the same over time. The mechanical energy includes the kinetic energy E k and the potential energy U . The individual energy in the initial and the final state changes, but the sum remains the same. This is only true when there are no non-conservative forces (such as friction) doing work. Back to Table 2 . 6