Introduction to Work

Where we have been Previously we used Newton’s Laws to analyze motion of objects Force and mass information were used to determine acceleration of an object (F=ma) We could use the acceleration to determine information about velocity or displacement Did the object speed up or slow down? How far did the object travel?

Where we are going Now we will take a new approach to looking at motion We will now look at work and power in relation to motion Today we will focus on “work”

Definition of “work” The everyday definition of “work” and the one that we use in physics are quite different from each other When most people think about “work”, they think of the job that they have Although it is possible that you are doing the physics definition of work while at your job, it is not always the case

Physics Definition of “Work” Like so many other things in physics, we have to use an exact definition to really explain what “work” is PHYSICS DEFINITION Work happens when a force causes an object to move through a displacement When a force acts upon an object to cause a displacement of the object, it is said that WORK has been done upon the object

Work There are three key ingredients to work Force Displacement Cause In order for a force to qualify as having done “work” on an object, there must be a displacement and the force must cause the displacement

Everyday Examples of “Work” There are several good examples of work which can be observed in everyday life A horse pulling a plow through a field A person pushing a shopping cart A student lifting a backpack onto her shoulder A weightlifter lifting a barbell above his head In each case described here there is a force exerted upon an object to cause that object to be displaced

Work Work – Exerting force in a way that makes a change in the world. Throwing a rock is work: you’re exerting a force, and the rock’s location changes (i.e. “the world has been changed”) Pushing on a brick wall is not work: you’re exerting a force, but “the world doesn’t change” (the wall’s position doesn’t change).

Work So exerting force alone isn’t enough. You have to both exert a force, and make a change. If you’re not exerting a force, you’re not doing work. Example: Throwing a ball. While you are “throwing the ball” (as opposed to just holding it) you are exerting a force on the ball. And the ball is moving. So you’re doing work. After the ball leaves your hand, you are no longer exerting force. The ball is still moving, but you’re no longer doing work.

(Work) = (Force exerted) (Displacement of object) (cos Θ) So, mathematically, we define work as “exerting a force that causes a displacement”: (Work) = (Force exerted) (Displacement of object) (cos Θ) or W = F*d*cosΘ W = Work done (J) F = Force exerted on object (N) d = Displacement of object (m) Θ = Angle between the force and the displacement

New Unit! The units for work are Nm (newton × meter). As we did with newton (which are kg m/s2), we will “define” the newton-meter to be a new unit. We’ll call this unit the joule. Abbreviation for joule: J So, 1 Nm = 1 J

Defining Θ – “the angle” This is a very specific angle Not just “any” angle - It is the angle between the force and the displacement Scenario A: A force acts rightward upon an object as it is displaced rightward. The force vector and the displacement vector are in the same direction, therefore the angle between F and d is 0 degrees Because the force and the displacement point in the same direction, there is nothing between them – NO angle - 0° “between” the vectors F Θ = 0 degrees d

Defining Θ – “the angle” Scenario B: A force acts upward upon an object as it is displaced rightward. The force vector and the displacement vector are at a right angle to each other, therefore the angle between F and d is 90 degrees F Θ = 90 degrees In order to pivot from F to d, you must rotate 90° d

Defining Θ – “the angle” Scenario D: A pushing force acts leftward upon an object as it is displaced up a ramp to the left. The force vector and the displacement vector need to be drawn starting from the same location in order to find the angle between them. d d F F Θ = 28 degrees In order to pivot from F to d, you must rotate 28°

Rules for Work If the force and displacement are in the same direction – the work done is POSITIVE because it adds energy If the force and displacement are in opposite directions – the work done is NEGATIVE because it removes energy If the force and displacement are perpendicular – the work done is equal to ZERO

To Do Work, Forces must CAUSE Displacement Consider scenario C from the previous slide The situation is similar to a waiter who carried a tray full of meals with one arm (F=20N) straight across a room (d=10m) at constant speed W = F*d*cosΘ W = (20N)(10m)(cos 90°) W = 0J The waiter does not do work upon the tray as he carries it across the room

The Meaning of Negative Work On occasion, a force acts upon a moving object to hinder a displacement A car skidding to a stop on a roadway surface A baseball player sliding to a stop on the infield dirt In such cases the force acts in the direction opposite the objects motion in order to slow it down The force doesn’t cause the displacement, but it opposes the displacement This results in negative work

Example of Work You are pushing a very heavy stone block (200 kg) across the floor. You are exerting 620 N of force on the stone, and push it a total distance of 20 m in 1 direction before you get tired and stop. How much work did you just do? W = (620 N)(20 m) = 12,400 J

Practice Problem Jessica Lee goes to the market and applies 200N of force at an angle of 35° above horizontal on the handle of the 600N shopping cart. She pushes the cart halfway down the freezer aisle which is 12 m long and stops to grab a bag of peas. She adds them to the cart which increases the amount of force she must push with by 15 N. She continues to the end of the freezer aisle. How much work did Jessica Lee do before she got the peas? 983.0 J How much work did Jessica Lee do total (in the freezer aisle)? 2039.7 J

Work Problems Austin lifts a 200 N box 4 meters. How much work did he do? W = (200N)(4m) W = 800 J

Work Problems Caitlin pushes and pushes on a loaded shopping cart for 2 hours with 100 N of force. The shopping cart does not move. How much work did Caitlin do? Chase lifts a 100 kg (220 lbs) barbell 2 meters. How much work did she do?

Work Done By “Lifting” Something Notice that when we were pushing something along the ground, the work done didn’t depend on the mass. Lifting up something does do work that depends on mass. Because of gravity: Gravity always pulls down with a force equal to m*ag, where m is the mass, and ag = -9.8 m/s2. So we must exert at least that much force to lift something. The more mass something has, the more work required to lift it.

Work Done By “Lifting” Something Example: A weightlifter lifts a barbell with a mass of 280 kg a total of 2 meters off the floor. What is the minimum amount of work the weightlifter did? The barbell is “pulled” down by gravity with a force of (280 kg)(9.8 m/s2) = 2,744 N So the weightlifter must exert at least 2,744 N of force to lift the barbell at all. --If that minimum force is used, the work done will be: W = (2,744 N)(2 m) = 5,488 J

We’ve talked about “work”, now let’s talk about “power”

“Work” Refresher Remember that the formula for finding the amount of work done upon an object is: W = (F)*(d) W = the work done F = the force required to cause displacement d = displacement of the object

Definition of Power Common definition usually includes something about strength Physics definition - the timed rate at which work is done depends directly upon the work and inversely upon the time to do that work If you increase work, then you increase power BUT If you increase time, then you decrease power

How Hard Are You Working? The rate at which work is done is called power: (Power) = (Work Done) / (Time Spent Working) P = W / t Power is “how hard” something is working.

Units for power: 1 J 1 s = 1 W one watt (W) equals one joule (J) per second (s) joule per second = watt

Power Example Let’s say that it took us 40 s to move that 200 kg stone block the 20 m. Remember that we did 12,400 J of work on the stone block Given Information: W = 12,400 J t = 40 s Basic Equation: P = W/t Working Equation: P = 12,400 J / 40 s P = 310 J/s Since it took us 40 s to move the block, we were doing 310 J of work per second OR generating 310 W of power

Fast work isn’t more work Go back to our 200 kg block example. Remember that when it took us 40 s to push the block the 20 m when we applied a force of 620 N, that implied that we had a power output of 310 W. If we exerted the same force (620 N) and pushed the block the same distance (20 m), but took half as long to do so (20 s), our power output would double to 620 W.

Fast work isn’t more work *But* notice that the total work done doesn’t change – we still exerted 620 N of force over a distance of 20 m. Copy this sentence in your notes: So increasing power output doesn’t mean you’re doing more work, it means you’re doing the work faster.

Energy comes in many forms: mechanical, electrical , magnetic, solar, What is energy?? Energy is the capacity of a physical system to perform work. Or the ability to make things move Energy is a scalar quantity…..remember what scalar means?? Energy comes in many forms: mechanical, electrical , magnetic, solar, thermal, chemical, etc...

How is energy divided? All Energy Potential Energy Kinetic Energy Gravitational Potential Energy Chemical Potential Energy Elastic Potential Energy

KE = 1/2 mv2 Kinetic Energy All moving objects that Energy of motion All moving objects that have mass have kinetic energy. KE = 1/2 mv2 m - mass of the object in kg v - speed of the object in m/s KE - the kinetic energy in J

Let’s practice a few problems…. Practice Problem 1: What is the kinetic energy of a 150 kg object that is moving with a speed of 15 m/s? First!!! What are you trying to solve for? KINETIC ENERGY Remember that KE = 1/2mv2 Write down your known variables: Mass is measured in kg so m = 150 kg Velocity is measured in m/s so v = 15 m/s Plug your numbers into the formula KE = ½ (150)(15)2 Solve your equation with units KE = 16875 J – Don’t forget, Energy is measured in Joules!

What are you trying to solve for? What’s the equation??? Practice Problem 2: What is the kinetic energy of a 0.05 kg bullet traveling at 475 m/s? What are you trying to solve for? KE What’s the equation??? KE = 1/2mv2 Write down your known variables: Mass is measured in kg so m = 0.05 kg Velocity is measured in m/s so v = 475 m/s Plug your numbers into the formula KE = ½ (0.05)(475)2 Solve your equation with units KE = 5640.63 J (round to the nearest hundredth place)

What are you trying to solve for? What’s the equation??? Practice Problem 3: A greyhound at a race track can run at a speed of 17.0 m/s. What is the KE of a 24 kg greyhound as it crosses the finish line? What are you trying to solve for? KE What’s the equation??? KE = 1/2mv2 Write down your known variables: Mass is measured in kg so m = 24 kg Velocity is measured in m/s so v = 17 m/s Plug your numbers into the formula KE = ½ (24)(17)2 Solve your equation with units KE = 3468 J (round to the nearest hundredth place)

Think about this… Two marbles, one twice as heavy as the other, are dropped to the ground from the roof of a building. Just before hitting the ground, the heavier marble has 1. as much kinetic energy as the lighter one. 2. twice as much kinetic energy as the lighter one. 3. half as much kinetic energy as the lighter one. 4. four times as much kinetic energy as the lighter one. 5. impossible to determine Think about the formula and write the question and the answer you think is right in your journal.

Potential Energy There are different ways that energy can be stored: Gravitational Potential Energy Elastic Potential Energy Chemical Potential Energy

Ug = mgh Potential Energy gravitational potential energy energy of position, location or condition gravitational potential energy Ug = mgh m - mass of object in kg g - acceleration of gravity in m/s2 h - height of object, in m PEg – gravitational potential energy in J

Let’s practice a few problems…. Practice Problem 1: Calculate the gravitational potential energy of a 1,500 kg elevator that is 45 meters above the ground floor. Choose the ground floor to be the reference level. First!!! What are you trying to solve for? POTENTIAL ENERGY Remember that PE = mgh Write down your known variables: Mass is measured in kg so m = 1500 kg Gravity is ALWAYS 9.8 m/s2 (this is the acceleration due to gravity on Earth) Height is measured in meters so h = 45 m Plug your numbers into the formula PE = (1500)(9.8)(45) Solve your equation with units PE = 661500 J – Don’t forget, Energy is measured in Joules!

1. What are you trying to solve for? POTENTIAL ENERGY Practice Problem 2: The crystal ball which drops on New Year’s Even in Times Square has a mass of 5,585 kg. If the ball descends 43 meters, what is its change in potential energy? 1. What are you trying to solve for? POTENTIAL ENERGY 2. What equation do you need to use to solve this? PE = mgh 3. Write down your known variables: Mass is measured in kg so m = 5585 kg Gravity is ALWAYS 9.8 m/s2 (this is the acceleration due to gravity on Earth) Height is measured in meters so h = 43 m 4. Plug your numbers into the formula PE = (5585)(9.8)(43) 5. Solve your equation with units PE = 2353519 J – Don’t forget, Energy is measured in Joules!

1. What are you trying to solve for? HEIGHT Practice Problem 3: Calculate the height above the Earth’s surface of a 5.5 kg object whose gravitational potential energy is equal to 5290 J. 1. What are you trying to solve for? HEIGHT 2. What equation do you need to use to solve this? PE = mgh 3. Write down your known variables: Mass is measured in kg so m = 55 kg Gravity is ALWAYS 9.8 m/s2 (this is the acceleration due to gravity on Earth) Gravitational PE is measured in Joules so PE = 5290 J 4. Notice you’ll have to re-arrange the formula to get height by itself. PE/mg = h 5. Plug your numbers into the formula h = 5290/(5.5*9.8) 6. h = 98.14 m – Don’t forget, height is measured in meters!

One more thing…. When you are calculating gravitational potential energy, your reference point is important. A zero height position has to be assigned Typically, the ground is considered to be a position of zero height, but not always The tabletop is usually assigned as the zero height position, since a lot of our labs are done on the tabletop This means if the potential energy is based upon the height of the tabletop. If the object goes beyond the tabletop, your height could be a negative value, which also makes your GPE value negative.

elastic potential energy: energy of position, location or condition elastic potential energy: energy stored in elastic materials as the result of their stretching or compressing Us = ½ kx2 k – elastic constant in N/m x - elongation or compression in m PEelastic – elastic potential energy in J Imagine the spring constant as how “stiff” or hard to pull the spring is.

Think about this… A spring-loaded toy dart gun is used to shoot a dart straight up in the air, and the dart reaches a maximum height of 24 m. The same dart is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the dart go this time, neglecting friction and assuming an ideal spring? 1. 96 m 2. 48 m 3. 24 m 4. 12 m 5. 6 m 6. 3 m 7. impossible to determine

W = ½ m v2 - ½ m v02 Work-Energy Theorem the net work done on an object is equal to its change in kinetic energy   W = ½ m v2 - ½ m v02