General Physics 1: Unit 7: Work, Power, and Energy PDF

Unit 7: Work, Power, and Energy Lesson 7.1 Work Contents Introduction 1 Learning Objectives 2 Warm Up 2 Learn about It! 3 Dot or Scalar Product...

Unit 7: Work, Power, and Energy Lesson 7.1 Work Contents Introduction 1 Learning Objectives 2 Warm Up 2 Learn about It! 3 Dot or Scalar Product 3 Work 11 Definition of Work in Physics 11 Work Done by a Constant Force 11 Work Done by a Varying Force 17 Key Points 18 Key Formulas 19 Check Your Understanding 20 Challenge Yourself 21 Bibliography 22 Key to Try It! 22 Unit 7: Work, Power, and Energy Lesson 7.1 Work Introduction In the last unit, you have learned about the different properties of forces. We know that even if we are just standing or sitting, there are still forces acting on us. Not all forces can cause a movement to a specific object. For example, even if you try to push a sturdy wall, the wall will not change its position. On the other hand, some forces can be used in moving objects like pushing the door to open, pulling your trolley bags, carrying books, and even just moving the objects up and down. In these situations, the forces that caused the change in the position of the object are said to be doing work on the object. Just like the image shown above, the man exerts force to the wheel, making the wheel move. Thus, the person is doing work on the wheel. In this lesson, we will further discuss the properties of forces and how they can produce work. 7.1. Work 1 Unit 7: Work, Power, and Energy Learning Objectives DepEd Competencies Calculate the dot or scalar product In this lesson, you should be able to do of vectors (STEM_GP12WE-If40). Determine the work done by a the following: force (not necessarily constant) Define the dot or scalar product. acting on a system (STEM_GP12WE-If41). Determine the work done by a Solve problems involving work, force acting on a system. energy, and power in contexts Solve problems that involve work such as, but not limited to, bungee jumping, design of roller-coasters, in real-life problems. number of people required to build structures such as the Great Pyramids and the rice terraces; power and energy requirements of human activities such as sleeping vs. sitting vs. standing, running vs. walking. (Conversion of joules to calories should be emphasized at this point.) (STEM_GP12WE-Ihi-55). Warm Up Positive, Negative, or Zero 10 minutes This activity will give you a hint on what will be discussed in this lesson. The materials needed and procedure for this is written below: Materials chair backpack 2 bottled water 7.1. Work 2 Unit 7: Work, Power, and Energy Procedure Group yourselves into groups of four. Each group should do the following: 1. Get a chair and push it for 1.0 meter. 2. Carry a backpack properly then walk at 1.0 meter. 3. Have two bottled water, make sure both hands are holding one of each. Then move the bottles up and down imagining you have a dumbbell on your hand. Move it for 15 seconds. 4. After doing the steps 1-3, the teacher will call one representative per group to answer the guide questions below. Guide Questions 1. Have you encountered situations similar to the procedures 1 to 3 in your everyday life? When? Describe briefly. 2. How will you describe the activity that you have done when it comes to its direction, displacement, and force exerted on the object? 3. With the activity that you have done, in what past lessons can you relate it to? Explain briefly why. Learn about It! There are a lot of available definitions for work. It could be the activity that we do every day and being paid for the effort of doing it. It could also be something that you exert some effort or time on. We may learn the definition of work in science, specifically in physics in this lesson, but before that, we need to review first about the dot or scalar product. Why is it important to know about dot or scalar product? Dot or Scalar Product The scalar product consists of two vectors, and (shown in Fig. 7.1.1) and it is written as. With this notation, the scalar product can also be called a dot product. Although 7.1. Work 3 Unit 7: Work, Power, and Energy and are vectors, the quantity is scalar. The scalar product can also be written and defined as: Fig. 7.1.1. Sample vectors and where A refers to the magnitude of , B refers to the magnitude of and θ refers to the angle made between the two vectors. The result of the scalar product of two vectors is the scalar quantity. How would you describe a scalar quantity? For vectors given by their components: and , the scalar product could be written as: Equation 7.1.1 7.1. Work 4 Unit 7: Work, Power, and Energy Remember Remember if the θ = 90°, then the cos(θ) = 0. Simply speaking, if the two vectors are perpendicular with each other if and only if their scalar product is equal to zero. Using the scalar product, you may also find the cosine and therefore the angle between two vectors using the following equation: Equation 7.1.2 The concept of calculating the dot or scalar product is what we use in calculating work. With this, we may now have an idea on how to determine the work done by a force that acts on a system. Let's Practice! Example 1 Calculate the dot product of and that is shown in the figure below: 7.1. Work 5 Unit 7: Work, Power, and Energy Solution Step 1: Identify what is required in the problem. You are asked to find the scalar product of the vectors given. Step 2: Identify the given in the problem. The magnitude of the is 10. The vector has a magnitude of 12. The angle between the two vectors is 50.0°. Step 3: Write the working equation. Step 4: Substitute the given values. Step 5: Find the answer. The dot product between vectors A and B is equal to 77.13. 1 Try It! Calculate the scalar product of = 11 and = 15 with an angle between them of 60.5º. Example 2 Find the scalar product between the and that is shown in the figure below: 7.1. Work 6 Unit 7: Work, Power, and Energy Solution Step 1: Identify what is required in the problem. You are asked to find the scalar product of the vectors given. Step 2: Identify the given in the problem. The magnitude of the is 18. The has a magnitude of 12. The angle between the x-axis and is 70.0°. The angle between the x-axis and is 138.0°. Step 3: Write the working equation. Before calculating the scalar product, the x- and y-components of the vectors should be determined first. Step 4: Substitute the given values. 7.1. Work 7 Unit 7: Work, Power, and Energy Step 5: Find the answer. The scalar product between vectors A and B is equal to 80.93. 2 Try It! Find the scalar product between and that is shown in the figure above. Example 3 Using the figure below find the following: a. The x- and y-components of and. b. The scalar product of the two vectors, where = 5 and =10. c. The θ between the two vectors. 7.1. Work 8 Unit 7: Work, Power, and Energy Solution Step 1: Identify what is required in the problem. You are asked to find the x- and y-components of the two vectors, scalar product, and the angle between the two vectors. Step 2: Identify the given in the problem. The magnitude of the is 5. The vector has a magnitude of 10. The angle between the x-axis and is 45.0°. The angle between the x-axis and is 111.0°. Step 3: Write the working equations. To solve for the x- and y-components, use the following: To solve for the scalar product, use the equation: To solve the angle between the two vectors, use: 7.1. Work 9 Unit 7: Work, Power, and Energy Step 4: Substitute the given values. For the x- and y-components of the vectors: For the scalar product of the two vectors: For the angle between the two vectors: Step 5: Find the answer. The components of the vectors are Ax= 3.54, Ay = 3.54 Bx = -3.58 By = 9.34, the scalar product is equal to 20.39 and the angle between them is 65.93°. 7.1. Work 10 Unit 7: Work, Power, and Energy 3 Try It! Using the figure below find the following: a. The x and y components of and. b. The scalar product of the two vectors, where = 10 and =20. c. The θ between the two vectors. Work Definition of Work in Physics In physics, work refers to the measure of energy transfer that occurs when an object is moved over a distance by an external force, at least part of which is applied in the direction of the displacement. In short, it is the force that causes the objects to move to result in its displacement. Remember Work is just a scalar quantity because it has no directions, it is just a magnitude and it can be positive or negative. Work Done by a Constant Force If the work is done in a constant force (constant in both magnitude and direction) it is defined as the product of the magnitude of the displacement multiplied by the component of the force parallel to the displacement. In an equation form, we can write it as: 7.1. Work 11 Unit 7: Work, Power, and Energy where F is the component of the constant force parallel to the displacement d. In general, we state the equation for work as: Equation 7.1.3 where F is the component of the constant force parallel to the displacement d and 𝜃 is the angle between the directions of the force and the displacement. We can deduce from Equation 7.1.3 that the work done by a force F is just the dot product between the magnitude of the force and the displacement of the object. Thus, work is scalar even if force and displacement are both vector quantities. See Figure 7.1.2 as reference. Figure 7.1.2. This figure shows a person pulling a box on a horizontal surface. The work done by the applied force F is , where d is the displacement and 𝜃 is the angle between the force and displacement. 7.1. Work 12 Unit 7: Work, Power, and Energy What is the importance of dot or scalar products in learning the concept of work? In calculating work, the SI unit that we use is the joule (J). In our formula, work is equal to force multiplied to displacement. The SI unit of force is newton (N) while the SI unit of displacement or distance is meter (m), meaning 1 joule is equivalent to 1 newton-meter (N m): 1 joule = (1 newton) (1 meter) or 1 J = 1 N m Did You Know? Joule is a unit of work or energy in the International System of Units (SI); it is equal to the work done by a force of one newton acting through one meter. This unit was named after the English physicist James Prescott Joule. This unit is equal to 107 ergs, or approximately 0.7377 foot-pounds. Does the direction of the force exerted on the system affect the value of work? There are some points that you need to remember for you to evaluate or determine if the work is positive, negative or zero. 1. If force is parallel to the direction of displacement, then work is positive. 2. If force is opposite to the direction of displacement, then work is negative. 3. If force is perpendicular to the direction of the displacement, then work is zero. 7.1. Work 13 Unit 7: Work, Power, and Energy Let's Practice! Example 4 Jenny and had her groceries done. Inside the store her cart was full of things that she bought. She needs to push her cart into the cashier area with a distance of 4.5 meters so she applied 50 newtons on her cart to make it through the cashier. Determine the work done by the force of Jenny acting on the cart. Solution Step 1: Identify what is asked in the problem. You are asked to determine the work done by the force acting on the system. Step 2: Identify the given in the problem. The displacement (d = 4.5 meters) and the force (F = 50 newtons) are both given. Step 3: Write the working equation. Step 4: Solve for work. Step 5: Calculate the final answer. The work done by the force acting on the system is 225 J. 4 Try It! A boy is playing with a box. He wants to pull the box 3.0 meters away from his position on a horizontal surface. For him to pull the box he exerted 20 newtons of force on the box. What is the work done by the force on the system? 7.1. Work 14 Unit 7: Work, Power, and Energy Example 5 A force F = 15 N acting on a box 1.5 m along a horizontal surface. The force acts at a 40.0o angle as shown in the figure below. Determine the work done by force F? Solution Step 1: Identify what is asked in the problem. You are asked to determine the work done by the force acting on the system. Step 2: Identify the given in the problem. The displacement (d = 1.5 meters), the force (F = 15 newtons), and the angle (𝜃 = 40.0o) are given. Step 3: Write the working equation. Step 4: Substitute the given to the formula. Step 5: Calculate the final answer. The work done by the force on the box is 17.24 J. 7.1. Work 15 Unit 7: Work, Power, and Energy 5 Try It! Kassandra has a new trolley bag that she used on her first day of class. She pulls her bag going to her classroom with a force of 180 N at an angle of 35° with a horizontal displacement of 25.0 m. Determine the work done by Kassandra on her trolley bag. Example 6 Determine the work done by Eric if the luggage bag with 10 kg is pulled for 45 m vertically with an angle of 35°. Solution Step 1: Identify what is asked in the problem. You are asked to determine the work done by the force acting on the system. Step 2: Identify the given in the problem. The displacement (d = 45 meters), the mass (m = 10 kg), and the angle (𝜃 = 35.0o) are given. Step 3: Write the working equation. Before calculating the work, force should be determined first using: Step 4: Substitute the given to the formula. 7.1. Work 16 Unit 7: Work, Power, and Energy Step 5: Calculate the final answer. The work done by the earth on the luggage bag is -3,612.46 J. 6 Try It! Herbert placed a ramp on his stairs following its 45o angle. He wants to push a box that is loaded with clothes to arrange his things in his room. The box is 15 kg and the distance of the stairs up to his room is 3.5 m. How much work is done by Herbert to the box? Work Done by a Varying Force In our everyday scenario, work does not always have constant force. Sometimes, we have a varying force acting on the system as the work occurs. For example when the rocket leaves the center of the Earth it needs more work to overcome the force of gravity. Another example is when we pull a heavy object going up the hill, we need to exert more force as we displace it for us to arrive at the desired destination. Also in springs, when we stretch the spring as it goes longer, we tend to exert more force on it. The work done by a varying force is determined graphically, so to do so we need to plot the force (F) as a function of distance (d). As you can see in Figure 7.1.3, the distance was divided into small segments Δd. In each segment indicated the average F in a horizontal dashed line. The work for each segment would be defined as W = FΔd, which is the area of the rectangle, where Δd is width and F is height. 7.1. Work 17 Unit 7: Work, Power, and Energy Figure 7.1.3. The graph of the work done by a varying force You can notice that the shaded part of the graph pertained to the total work done by the force. Thus, we can say that work done by a certain for F is the area under the force vs. distance graph. Key Points ___________________________________________________________________________________________ The concept of dot or scalar product is used to determine the work done by a force on a system. Work is positive if the F and d is parallel, negative if it is opposite in direction, and there is no work if F and d is perpendicular. Work can be done by: ○ Constant Force ○ Varying Force The area under the force vs. distance graph pertains to the total work done. ___________________________________________________________________________________________ 7.1. Work 18 Unit 7: Work, Power, and Energy Key Formulas ___________________________________________________________________________________________ Concept Formula Description Dot or Scalar Use this formula to solve for Product scalar products if the where: magnitude of the two A is the magnitude of vectors are given and the vector angle between them. B is the magnitude of vector θangle between the two vectors Dot or Scalar Use this formula to solve for Product scalar products if the x and y-component of the two where: vectors are given. Ax is x-component of Ay is y-component of Bx is x-component of By is y-component of Work Use this formula to solve for work if there is an angle where: between the force and W is work in joules (J) distance. F is force in newtons (N) d is distance in meters (m) 𝜃 is the angle between the force and displacement ___________________________________________________________________________________________ 7.1. Work 19 Unit 7: Work, Power, and Energy Check Your Understanding A. Identify the following. Write the correct answer on the space provided before the number. ______________________ 1. It is the result of dot product. ______________________ 2. It occurs when we exert effort on an object for it to move. ______________________ 3. It would be the sign of work if the displacement and force are of the same direction. ______________________ 4. It would be the value of work if a person exerted a force to the object but the object moved perpendicular to the force. ______________________ 5. It is the SI unit of work? B. Modified True or False. Write T if the statement is true. If false, change the underlined word to make the statement true. ______________________ 1. The SI unit for work is equal to 1 Newton-meter. ______________________ 2. Work is a vector quantity. ______________________ 3. If the F and d is perpendicular to each other work is negative. ______________________ 4. There is no change in displacement when the work is done by a constant force. ______________________ 5. There is a change in displacement when the work is done by varying force. ______________________ 6. Acceleration is the dot product of force and the displacement of the object. ______________________ 7. If we just hold an object there is no work done. ______________________ 8. Work could be positive or negative. 7.1. Work 20 Unit 7: Work, Power, and Energy ______________________ 9. Work is a vector quantity. ______________________ 10. We may determine the work done by varying forces graphically. C. Solve the following problems. 1. What would be the scalar product of the vectors A and B if they have an angle of 50.0o? Each vector has a magnitude of A = 15 and B = 20. 2. What is the dot product of = 7 and = 12? The angle from x-axis to is 45.0o while the angle from x-axis to is 103.0o. 3. Renee and his friends were riding in a car and suddenly it stopped and had a hard time to start. For them to proceed to the car repair shop they need to push the car with a force of 1080 N with a distance of 100 m. Determine the work done by Renee and his friends in the car. 4. A little girl holds a 5 N glass and walks a certain distance of 3 m. What is the work done by the girl to the glass? 5. Katherine is doing her work-out. She is lifting her 5-kg dumbbell for 0.80 m. How much work is done by Katherine on her dumbbell? Challenge Yourself Answer what is being asked on the following: 1. Compare the work being done when you are carrying your bag versus when you are pulling your bag. 2. Assess what will happen to the work done if you are going up on a hill and pulling a sack of stones. 3. Explain the factors that make the work positive, negative or zero. 4. Explain why we need to go understand the dot or scalar product before learning the 7.1. Work 21 Unit 7: Work, Power, and Energy concept of work. 5. Relate the concept of work in real life situations. Bibliography Giancoli, Douglas C. Physics: Principles with Applications (version 7th edition). Upper Saddle River, NJ: Pearson Education Inc., 2014. Knight, Randall Dewey, and Randall Dewey. Knight. Student Workbook: Physics for Scientists and Engineers, a Strategic Approach with Modern Physics, 4/E. Boston: Pearson, 2017. Silverio, Angelina A., and Gloria de Castro. Bernas. Physics: Exploring Life through Science. Quezon City: Phoenix Publishing House, Inc., 2012. Young, Hugh D., Roger A. Freedman, A. Lewis Ford, and Hugh D. Young. University Physics. Boston, MA: Pearson, 2014. Key to Try It! 1. 81.25 2. 46.54 3. a. Ax = 6.23, Ay = 7.83, Bx = -9.23, By = 17.74 b. 81.40 c. 65.98° 4. 60 J 5. 0.82 J 6. 363.81 J 7.1. Work 22