Work Energy Power Whenever work is being done

Work, Energy, Power

• Whenever work is being done, energy is being changed from one form to another or being transferred from one body to another • The amount of work done on a system is the change in energy of the system W=∆E • Work tells us how much a force or combination of forces changes the energy of a system

WORK • When a force acts upon an object to cause a displacement of the object, it is said that work was done upon the object. • There are three key ingredients to work - force, displacement, and 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.

Was Work Done? • A teacher applies a force to a wall and becomes exhausted. • No. This is not an example of work. The wall is not displaced. A force must cause a displacement in order for work to be done. • A book falls off a table and free falls to the ground. • Yes. This is an example of work. There is a force (gravity) which acts on the book which causes it to be displaced in a downward direction (i. e. , "fall"). • A waiter carries a tray full of meals above his head by one arm straight across the room at constant speed. • No. This is not an example of work. There is a force (the waiter pushes up on the tray) and there is a displacement (the tray is moved horizontally across the room). Yet the force does not cause the displacement. To cause a displacement, there must be a component of force in the

Work Equation • Mathematically, work can be expressed by the following equation: W = F • d • cos Θ or W=F d • F is the force, d is the displacement, and the angle (theta) is defined as the angle between the force and the displacement vector.

(only write what is in bold) • It was mentioned earlier that the waiter does not do work upon the tray as he carries it across the room. The force supplied by the waiter on the tray is an upward force and the displacement of the tray is a horizontal displacement. • The angle between the force and the displacement is 90 degrees. • Regardless of the magnitude of the force and displacement, F*d*cosine 90 degrees is 0 (since the cosine of 90 degrees is 0). • A vertical force can never cause a horizontal displacement; thus, a vertical force does not do work on a horizontally displaced object!!

• Ɵ is the angle between the force and the displacement vector • A force is applied to a cart to displace it up the incline at constant speed. Several incline angles are typically used; yet, the force is always applied parallel to the incline. The displacement of the cart is also parallel to the incline. Since F and d are in the same direction, the angle theta in the work equation is 0 degrees.

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units • Work (Joules) Force (Newton) Displacement (Meters) • Joule = N m or kg m 2/s 2 • One Joule of work is done when 1 N of force acts on a body moving it 1 m

• Find the amount of work done:

• Diagram A Answer: • W = (100 N) * (5 m)* cos(0 degrees) = 500 J The force and the displacement are given in the problem statement. It is said (or shown or implied) that the force and the displacement are both rightward. Since F and d are in the same direction, the angle is 0 degrees. • Diagram B Answer: • W = (100 N) * (5 m) * cos(30 degrees) = 433 J The force and the displacement are given in theproblem statement. It is said that the displacement is rightward. It is shown that the force is 30 degrees above the horizontal. Thus, the angle between F and d is 30 degrees. • Diagram C Answer: • W = (147 N) * (5 m) * cos(0 degrees) = 735 J The displacement is given in the problem statement. The applied force must be 147 N since the 15 -kg mass (Fgrav=147 N) is lifted at constant speed. Since F and d are in the same direction, the angle is 0 degrees.

• A 10 -N force is applied to push a block across a friction free surface for a displacement of 5. 0 m to the right. • What are the forces doing the work on the object and the amount of work done?

• Only Fapp does work. Fgrav and Fnorm do not do work since a vertical force cannot cause a horizontal displacement • Wapp= (10 N) * (5 m) *cos (0 degrees) = +50 Joules

• A 10 -N frictional force slows a moving block to a stop after a displacement of 5. 0 m to the right. • What are the forces doing the work on the object and the amount of work done?

• Only Ffrict does work. Fgrav and Fnorm do not do work since a vertical force cannot cause a horizontal displacement. • Wfrict =(10 N) * (5 m) * cos (180 degrees) = -50 Joules

• A 10 -N force is applied to push a block across a frictional surface at constant speed for a displacement of 5. 0 m to the right. • What are the forces doing the work on the object and the amount of work done?

• Fapp and Ffrict do work. Fgrav and Fnorm do not do work since a vertical force cannot cause a horizontal displacement • Wapp = (10 N) * (5 m) * cos (0 deg) = +50 Joules • Wfrict = (10 N) * (5 m) * cos (180 deg) = -50 Joules

• An approximately 2 -kg object is sliding at constant speed across a friction free surface for a displacement of 5 m to the right. • What are the forces doing the work on the object and the amount of work done?

• Neither of these forces do work. Forces do not do work when they makes a 90 -degree angle with the displacement. • No work is done

• An approximately 2 -kg object is pulled upward at constant speed by a 20 -N force for a vertical displacement of 5 m.

• Fgrav and Ftens do work. Forces do work when there is some component of force in the same or opposite direction of the displacement • Wtens = (20 N) * (5 m) * cos (0 deg) = +100 Joules • Wgrav = (20 N) * (5 m) * cos (180 deg) = -100 Joules

• Tim pushes a cart 10 kg cart across the floor at a constant velocity of 3 m/s. The coefficient of friction between the cart and floor is. 5 • A)How much work does Tim do if he pushed the box for 15 m? • B)How much work does Friction do? • C) What is the amount of the New Work?

• A) Fnet = 0 so Fapp = Ffrict= u. Fnet = (. 5)(10 kg)(9. 81 m/s 2) = 49. 05 N W=F*d=(49. 05 N)(15 m)=735. 75 J • B)-735. 75 J • C) 0 (zero net work = constant speed or no motion)

Solve Graphically • The work done by a force is the area enclosed under the curve

• How much work was done to move the object 30 m?

• Find the area of the rectangle (Lx. W) and the area of the triangle (1/2 BH) and add them together for the total work done. • W=(100 N)(20 m) + ½(10 m)(100 N) = 2500 J

• How much work was done to move the object from 20 to 25 m?

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Power • https: //www. youtube. com/watch? v=Z 7 z. QK 6 IV 2 vk

POWER

• The quantity that has to do with the rate at which a certain amount of work is done is known as the power. • Power is the rate at which work is done. It is the work/time ratio • Power = Work / time or P=W/t

• The standard metric unit of power is the Watt • A Watt is equivalent to a Joule/second • For historical reasons, the horsepower is occasionally used to describe the power delivered by a machine. One horsepower is equivalent to approximately 750 Watts. • The power equation suggests that a more powerful engine can do the same amount of work in less time.

Question • 1. Two physics students, Will N. Andable and Ben Pumpiniron, are in the weightlifting room. Will lifts the 100 -pound barbell over his head 10 times in one minute; Ben lifts the 100 -pound barbell over his head 10 times in 10 seconds. Which student does the most work? _______ Which student delivers the most power? _______ Explain your answers. • Ben and Will do the same amount of work. They apply the same force to lift the same barbell the same distance above their heads. • Yet, Ben is the most "power-full" since he does the same work in less time. Power and time are inversely proportional.

Question • During a physics lab, Jack and Jill ran up a hill. Jack is twice as massive as Jill; yet Jill ascends the same distance in half the time. Who did the most work? _______ Who delivered the most power? _______ Explain your answers. • Jack does more work than Jill. Jack must apply twice the force to lift his twice-as-massive body up the same flight of stairs. Yet, Jill is just as "power-full" as Jack. Jill does one-half the work yet does it one-half the time. The reduction in work done is compensated for by the reduction in time.

Question • When doing a chin-up, a physics student lifts her 42. 0 -kg body a distance of 0. 25 meters in 2 seconds. What is the power delivered by the student's biceps? • to raise her body upward at a constant speed, the student must apply a force which is equal to her weight (m • g). The work done to lift her body is • W = F * d = (411. 6 N) * (0. 250 m) W = 102. 9 J • The power is the work/time ratio which is (102. 9 J) / (2 seconds) = 51. 5 Watts (rounded)

Energy

• Potential energy is the stored energy of position possessed by an object.

Gravitational Potential Energy • Gravitational potential energy is the energy stored in an object as the result of its vertical position or height • dependent on two variables - the mass of the object and the height to which it is raised • There is a direct relation between gravitational potential energy and the mass of an object. More massive objects have greater gravitational potential energy. • There is also a direct relation between gravitational potential energy and the height of an object. The higher that an object is elevated, the greater the gravitational potential energy.

Gravitational Potential Energy PEgrav = m * • g • h In the above equation, m represents the mass of the object, h represents the height of the object and g represents the gravitational field strength (9. 8 N/kg on Earth)

• To determine the gravitational potential energy of an object, a zero height position must first be arbitrarily assigned (could be the ground or table top) • Since the gravitational potential energy of an object is directly proportional to its height above the zero position, a doubling of the height will result in a doubling of the gravitational potential energy. A tripling of the height will result in a tripling of the gravitational potential energy.

• Knowing that the potential energy at the top of the tall platform is 50 J, what is the potential energy at the other positions shown on the stair steps and the incline?

• A: PE = 40 J (since the same mass is elevated to 4/5 -ths height of the top stair) • B: PE = 30 J (since the same mass is elevated to 3/5 -ths height of the top stair) • C: PE = 20 J (since the same mass is elevated to 2/5 -ths height of the top stair) • D: PE = 10 J (since the same mass is elevated to 1/5 -ths height of the top stair) • E and F: PE = 0 J (since the same mass is at the same zero height position as shown for the bottom stair).

• A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat -top. If the mass of the loaded cart is 3. 0 kg and the height of the seat top is 0. 45 meters, then what is the potential energy of the loaded cart at the height of the seat-top? • If a force of 14. 7 N is used to drag the loaded cart (from previous question) along the incline for a distance of 0. 90 meters, then how much work is done on the loaded cart?

• PE = m*g*h • PE = (3 kg ) * (9. 8 m/s/s) * (0. 45 m) • PE = 13. 2 J • W = F * d * cos Theta • W = 14. 7 N * 0. 9 m * cos (0 degrees) • W = 13. 2 J

Elastic potential energy • Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. • Elastic potential energy can be stored in rubber bands, bungee chords, trampolines, springs, an arrow drawn into a bow, etc. • The amount of elastic potential energy stored in such a device is related to the amount of stretch of the device the more stretch, the more stored energy.

springs • Springs are a special instance of a device that can store elastic potential energy due to either compression or stretching. • A force is required to compress a spring; the more compression there is, the more force that is required to compress it further. • For certain springs, the amount of force is directly proportional to the amount of stretch or compression (x); the constant of proportionality is known as the spring constant (k). • Fspring = k • x

The slope of the line = K (the higher the K value the stiffer the spring) The area under the curve = the amount of work done in stretching the spring and PE stored in the spring

Springs • Such springs are said to follow Hooke's Law. If a spring is not stretched or compressed, then there is no elastic potential energy stored in it. • The equilibrium position is the position that the spring naturally assumes when there is no force applied to it. (zero potential energy) • There is a special equation for springs that relates the amount of elastic potential energy to the amount of stretch (or compression) and the spring constant. The equation is: • PEspring = ½ k x 2

Summary • potential energy is the energy that is stored in an object due to its position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height. An object possesses elastic potential energy if it is at a position on an elastic medium other than the equilibrium position.

Question • A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat -top. If the mass of the loaded cart is 3. 0 kg and the height of the seat top is 0. 45 meters, then what is the potential energy of the loaded cart at the height of the seat-top? • PE = m*g*h PE = (3 kg ) * (9. 8 m/s/s) * (0. 45 m) • PE = 13. 2 J

Kinetic Energy • Kinetic energy is the energy of motion. An object that has motion - whether it is vertical or horizontal motion - has kinetic energy. • The amount of kinetic energy that an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. The following equation is used to represent the kinetic energy (KE) of an object: • KE = 0. 5 • m • v 2

Kinetic Energy • This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. • That means that for a twofold increase in speed, the kinetic energy will increase by a factor of four. For a threefold increase in speed, the kinetic energy will increase by a factor of nine and for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen. • The kinetic energy is dependent upon the square of the speed. As it is often said, an equation is not merely a recipe for algebraic problem solving, but also a guide to thinking about the relationship between quantities.

Kinetic Energy • Kinetic energy is a scalar quantity; it does not have a direction. • Like work and potential energy, the standard metric unit of measurement for kinetic energy is the Joule. •

Question • Determine the kinetic energy of a 625 -kg roller coaster car that is moving with a speed of 18. 3 m/s. • KE = 0. 5*m*v 2 KE = (0. 5) * (625 kg) * (18. 3 m/s)2 • KE = 1. 05 x 105 Joules

Question • Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12 000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then what is her speed? • KE = 0. 5*m*v 2 12 000 J = (0. 5) * (40 kg) * v 2 • 300 J = (0. 5) * v 2 • 600 J = v 2 • v = 24. 5 m/s

Work Energy Theorem • The net work due to all forces equals the change in energy of a system: • Wnet = ∆KE Wnet = ∆PE • Work causes energy change and energy change causes work

• A 200 kg train is moving at a speed of 20 m/s. How much force must the breaks apply in order to stop the train in 10 m? • Wnet = ∆KEfinal - ∆KEinitial

Conservation of energy • When work is done upon an object, that object gains energy. The energy acquired by the objects upon which work is done is known as mechanical energy. • An object that possesses mechanical energy is able to do work • Mechanical energy can be either kinetic energy (energy of motion) or potential energy (stored energy of position). • The total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy

• Certain types of forces, that when present change the total mechanical energy of the object. Other types of forces can never change the total mechanical energy of an object, but rather can only transform the energy of an object from potential energy to kinetic energy (or vice versa).

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