Bell Ringer: Define Energy. Define Kinetic Energy. Define Potential Energy. Define Work.

Notes 5.1: Work and Kinetic Energy An Adventure created by Billy J. Jenkins

Objectives: Calculate Work. Determine the net work being done on an object.

Vocabulary: Work Net Work Energy Kinetic Energy joule

Active Physics Reference: Chapter 3 – Section 3: p. 279 – p. 287 (Energy and Work)

Further Learning: Physics: Principles and Problems Chapter 10 – Section 1: p. 257 – p. 265 (Energy and Work)

Video: https://www.youtube.com/watch?v=w4QFJb9a8vo

Energy: Energy can be transformed from one type to another and transferred from one object to another, but the total amount is always the same. Energy is conserved.   Energy is a scalar quantity (not a vector).

Kinetic Energy (K):   Kinetic energy – the energy associated with the state of motion of an object. In other words, the kinetic energy of an object is the energy that the object possesses due to its motion. An object that has motion – has kinetic energy. Kinetic energy is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. 𝐾= 1 2 𝑚 𝑣 2

S.I. Unit for Energy is: 1 𝑗𝑜𝑢𝑙𝑒=1 𝐽=1 𝑘𝑔∙ 𝑚 2 𝑠 2

CHECKPOINT I: What does it mean to say that Energy is Conserved? What is Kinetic Energy? What is the equation for Kinetic Energy? What is the SI Unit for Energy (Kinetic Energy, Potential Energy, Work, or Heat)?

Work (W): Work W is the energy transferred to or from an object by means of a force acting on the object . W = Fd Energy transferred to the object is positive work Energy transferred from the object is negative work. EMPHASIS: Work is transferred energy; doing work is the act of transferring the energy.

Very Important: Work is a scalar quantity. To calculate the work a force does on an object as the object moves through some displacement, we use only the force component along the object’s displacement. The force component perpendicular to the displacement does zero work. Work Done by a Constant Force: 𝑊=𝐹𝑑 cos 𝜃 F = the magnitude of force F d = displacement 𝜃 = the angle between the directions of the displacement d and the force F

Two Restrictions to the use of the equation W = Fd and W = Fdcosθ:   The force must be a constant force i.e. must not change in magnitude or direction as the object moves. 2) The object must be particle-like i.e. the object is rigid – that is, all parts of it move together in the same direction.

The SI unit for work is the same as the SI unit for energy which is the Joule since work is the transfer of energy.

Problem-Solving Strategies: WORK Sketch the system and show the force that is doing the work. Draw the force and displacement vectors of the system. Find the angle between each force and displacement. Calculate the work done by each force using 𝑊=𝐹𝑑𝑐𝑜𝑠𝜃. Calculate the net work done. Check the sign of the work using the direction of energy transfer. If the energy of the system has increased, the work done by that force is positive. If the energy has decreased, then the work done by that force is negative.

Example Problem 1: Work and Energy A 105-g hockey puck is sliding across the ice. A player exerts a constant 4.50 N force over a distance of 0.150 m. How much work does the player do on the puck? Step 1: Sketch the situation showing initial conditions, including a vector diagram.

Example Problem 1: Work and Energy A 105-g hockey puck is sliding across the ice. A player exerts a constant 4.50 N force over a distance of 0.150 m. How much work does the player do on the puck? Step 2: Write down your knowns and your unknowns. Knowns: Unknowns: m = 105 g = 0.105 kg Work = ? F = 4.50 N d = 0.150 m 𝜃=0°

(We start with our equation) Example Problem 1: Work and Energy A 105-g hockey puck is sliding across the ice. A player exerts a constant 4.50 N force over a distance of 0.150 m. How much work does the player do on the puck? Step 3: Solve for the unknowns. (We start with our equation) 𝑊=𝐹𝑑 cos 𝜃 (We plug in any zero’s) 𝑊=𝐹𝑑 cos (0) (In this case, we know cos(0) = 1, so we simply have the equation below) W = Fd

Example Problem 1: Work and Energy A 105-g hockey puck is sliding across the ice. A player exerts a constant 4.50 N force over a distance of 0.150 m. How much work does the player do on the puck? Step 4: Plug and Chug W = Fd W = (4.50 N)(0.150 m) W = 0.675 J

Example Problem 1: Work and Energy A 105-g hockey puck is sliding across the ice. A player exerts a constant 4.50 N force over a distance of 0.150 m. How much work does the player do on the puck? Step 5: Evaluate the answer. Are the units correct? Work is measured in joules. Does the sign make sense The player (external world) does work on the puck (the system). So the sign of work should be positive.

CHECK POINT 2: What is Work? What is equation for Work? Work done on an object is what kind of work? Positive or Negative? Work done by an object is what kind of work? Positive or Negative?

(Net Work):   When two or more forces act on an object, the net work done on the object is the sum of the works done by the individual forces. The net work can be found in two ways:   Find the work done by each force and then sum those works to find the net work 𝑊 𝑛𝑒𝑡 . W net = W 1 + W 2 +…+ W N = F 1 d 1 + F 2 d 2 +…+ F N d N

2) First, find the net force F net of those forces and substitute the found net force F net into:   W=Fdcos𝜙= F net dcos𝜙 The angle is the angle between the directions of the net force F net and the displacement vector 𝑑 for 𝑊= F net ∙ 𝑑 .

Example 2: Net Work (video) https://www.youtube.com/watch?v=udgMh3Y-dTk

Example 2: Net Work (Trashcan Problem – using Net Force Method) You have to move a gross trashcan with a mass of 4 kg a distance of 10 meters. You decide the best method to keep yourself clean is to tie a rope around it and pull it. You pull with a force of 50 N to which a 30 N frictional force opposes your motion. Find the net work done on the trash can. Step 1: Find your Net Force. There are four forces acting on the trashcan: Applied force, frictional force, gravitational force, and normal force. In this instance, we do not have to concern ourselves with any forces on the y-axis meaning we can ignore our normal force and gravitational force.

Example 2: Net Work (Trashcan Problem – using Net Force Method) You have to move a gross trashcan with a mass of 4 kg a distance of 10 meters. You decide the best method to keep yourself clean is to tie a rope around it and pull it. You pull with a force of 50 N to which a 30 N frictional force opposes your motion. Find the net work done on the trash can. Step 1: Find your Net Force. Since our direction of motion is solely on the x-axis, we are only concerned with those two forces. Therefore, our net force is: 𝐹 𝑛𝑒𝑡 = 𝐹 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 − 𝑓 𝑘 𝐹 𝑛𝑒𝑡 =50 𝑁−30 𝑁=20 N

(Remember that cos(0) = 1) Example 2: Net Work (Trashcan Problem – using Net Force Method) You have to move a gross trashcan with a mass of 4 kg a distance of 10 meters. You decide the best method to keep yourself clean is to tie a rope around it and pull it. You pull with a force of 50 N to which a 30 N frictional force opposes your motion. Find the net work done on the trash can. Step 2: Find the angle. 𝑊= 𝐹 𝑛𝑒𝑡 𝑑𝑐𝑜𝑠𝜃 (The angle is the angle between the displacement and the net force, which is 0 in this case) 𝑊= 𝐹 𝑛𝑒𝑡 𝑑𝑐𝑜𝑠 0 = 𝐹 𝑛𝑒𝑡 𝑑 (Remember that cos(0) = 1)

Example 2: Net Work (Trashcan Problem – using Net Force Method) You have to move a gross trashcan with a mass of 4 kg a distance of 10 meters. You decide the best method to keep yourself clean is to tie a rope around it and pull it. You pull with a force of 50 N to which a 30 N frictional force opposes your motion. Find the net work done on the trash can. Step 3: Find the net work. 𝑊= 𝐹 𝑛𝑒𝑡 𝑑 𝑊= 20 𝑁 10 𝑚 W = 200 J

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Exit Ticket: Determine the work done in the following situations. Jim Neysweeper is applying a 21.6-N force downward at an angle of 57.2° with the horizontal to displace a broom a distance of 6.28 m. (b) Ben Pumpiniron applies an upward force to lift a 129-kg barbell to a height of 1.98 m at a constant speed. (c) An elevator lifts 12 occupants up 21 floors (76.8 meters) at a constant speed. The average mass of the occupants is 62.8 kg. 73.5 J 2503 J 5.67•105 J