Work & Energy Work & Energy Discussion Discussion Definition Definition Dot Product Dot Product Work of a constant force Work of a constant force Work/kinetic energy theorem Work/kinetic energy theorem Work of multiple constant forces Work of multiple constant forces Comments Comments SPH3U: Introduction to Work and Energy

Work & Energy One of the most important concepts in physics One of the most important concepts in physics Alternative approach to mechanics Alternative approach to mechanics Many applications beyond mechanics Many applications beyond mechanics Thermodynamics (movement of heat) Thermodynamics (movement of heat) Quantum mechanics... Quantum mechanics... Very useful tools Very useful tools You will learn new (sometimes much easier) ways to solve problems You will learn new (sometimes much easier) ways to solve problems

Energy is.. The ability to do work Measured in Joules We use energy to do whatever task we need to do (move a car, lift a pencil, etc)

Types of Energy Energy can come in many forms including: – Thermal (energy in the form of moving atoms) – Electrical (energy possessed by charged particles) – Radiant (energy found in EM waves) – Nuclear (energy stored in holding the atom together) – Gravitational (energy stored due to a raised elevation)

Types of Energy cont. – Kinetic (energy due to motion of objects) – Elastic (energy stored in compression or stretch) – Sound (energy in vibrations) – Chemical (energy stored in molecular bonds)

Energy 101 Energy cannot be created or destroyed, it can only transform from one form to another. Eg. A student turns on the stove to heat a pot of water Electrical  Radiant  Thermal

Energy and Work We know energy is the ability to do work, but what is work? In physics work is the energy transferred to an object by an applied force over a displacement

Forms of Energy Kinetic: Energy of motion. Kinetic: Energy of motion. A car on the highway has kinetic energy. A car on the highway has kinetic energy. We have to remove this energy to stop it. We have to remove this energy to stop it. The brakes of a car get HOT! The brakes of a car get HOT! This is an example of turning one form of energy into another (thermal energy). This is an example of turning one form of energy into another (thermal energy).

Mass = Energy Particle Physics: Particle Physics: + 5,000,000,000 V e- - 5,000,000,000 V e+ (a) (b) (c) E = eV M E = MC 2 ( poof ! )

Energy Conservation Energy cannot be destroyed or created. Energy cannot be destroyed or created. Just changed from one form to another. Just changed from one form to another. We say energy is conserved! We say energy is conserved! True for any closed system. True for any closed system. i.e. when we put on the brakes, the kinetic energy of the car is turned into heat using friction in the brakes. The total energy of the “car-brakes-road-atmosphere” system is the same. i.e. when we put on the brakes, the kinetic energy of the car is turned into heat using friction in the brakes. The total energy of the “car-brakes-road-atmosphere” system is the same. The energy of the car “alone” is not conserved... The energy of the car “alone” is not conserved... It is reduced by the braking. It is reduced by the braking. Doing “work” on an isolated system will change its “energy”... Doing “work” on an isolated system will change its “energy”...

Work Mechanical work is done on an object when a force displaces an object. Note that this equation only applies when the force is constant and the force and displacement are in the same direction. When the force and displacement are not entirely in the same direction, the component of the force in the direction of the displacement is used. W=FΔd

Definition of Work: Ingredients: Fr Ingredients: Force (F), displacement (  r) F Work, W, of a constant force F r acting through a displacement  r is: F  rrr W = F   r = F  r cos  = F r  r  F rrrr displacement FrFr “Dot Product” The dot product allows us to multiply two vectors, but just the components that are going in the same direction (usually along the second vector)

Definition of Work... Only the component of F along the displacement is doing work. Only the component of F along the displacement is doing work. Example: Train on a track. Example: Train on a track. F r rr r  F cos 

Aside: Dot Product (or Scalar Product) Definition: ab a. b= ab cos  = a[b cos  ] = ab a = b[a cos  ] = ba b Some properties: a  bb  a a  b= b  a a  bb  a b  a q(a  b) = (qb)  a = b  (qa) (q is a scalar) a  b ca  b a  c c a  (b + c) = (a  b) + (a  c) (c is a vector) The dot product of perpendicular vectors is 0 !! a abab b ab baba

Work: Example (constant force) A force F = 10 N pushes a box across a frictionless floor for a distance  x = 5 m. A force F = 10 N pushes a box across a frictionless floor for a distance  x = 5 m. xxxx F byF on Work done by F on box is: F  xFx W F = F   x = F  x (since F is parallel to  x) Joules (J) W F = (10 N) x (5 m) = 50 Joules (J)

Units of Work: Force x Distance = Work Force x Distance = Work N-m (Joule) Dyne-cm (erg) = J BTU= 1054 J calorie= J foot-lb= J eV= 1.6x J cgsothermks Newton x [M][L] / [T] 2 Meter = Joule [L] [M][L] 2 / [T] 2

Example 1 How much mechanical work will be done pushing a shopping cart 3.5m with a force of 25N in the same direction as the displacement?

Example 2 A curler applies a force of 15.0N on a curling stone and accelerates the stone from rest to a speed of 8.00m/s in 3.5s. Assuming the ice surface is level and frictionless, how much mechanical work is done on the stone?

Example 3 Calculate the work done by a custodian on a vacuum cleaner if the custodian exerts an applied force of 50.0N on the vacuum hose and the hose makes a 30° angle with the floor. The vacuum moves 3.0m to the right on a flat level surface.

Useless work? No work: when there is no displacement, no work is done! Work can also be positive or negative relative to the motion, as shown in the next example.

Comments: Time interval not relevant Time interval not relevant Run up the stairs quickly or slowly...same Work Run up the stairs quickly or slowly...same Work Since W = F   r No work is done if: No work is done if: F = 0 or F = 0 or  r = 0 or  r = 0 or  = 90 o  = 90 o

Example 4 A shopper pushes a shopping cart on a horizontal surface with a horizontal applied force of 41.0N for 11.0m. The cart experiences a force of friction of 35.0N. Calculate the total work done on the cart.

Work & Kinetic Energy: A force F = 10 N pushes a box across a frictionless floor for a distance  x = 5 m. The speed of the box is v 1 before the push and v 2 after the push. A force F = 10 N pushes a box across a frictionless floor for a distance  x = 5 m. The speed of the box is v 1 before the push and v 2 after the push. xxxx F v1v1 v2v2 m

Work & Kinetic Energy... Since the force F is constant, acceleration a will be constant. We have shown that for constant a: Since the force F is constant, acceleration a will be constant. We have shown that for constant a: v v 1 2 = 2a(x 2 -x 1 ) = 2a  x. v v 1 2 = 2a(x 2 -x 1 ) = 2a  x. multiply by 1 / 2 m: 1 / 2 mv / 2 mv 1 2 = ma  x multiply by 1 / 2 m: 1 / 2 mv / 2 mv 1 2 = ma  x But F = ma 1 / 2 mv / 2 mv 1 2 = F  x But F = ma 1 / 2 mv / 2 mv 1 2 = F  x xxxx F v1v1 v2v2am

Work & Kinetic Energy... So we find that So we find that 1 / 2 mv / 2 mv 1 2 = F  x = W F 1 / 2 mv / 2 mv 1 2 = F  x = W F Define Kinetic Energy K: K = 1 / 2 mv 2 Define Kinetic Energy K: K = 1 / 2 mv 2 K 2 - K 1 = W F K 2 - K 1 = W F W F =  K (Work/kinetic energy theorem) W F =  K (Work/kinetic energy theorem) xxxx F am v2v2 v1v1

Work/Kinetic Energy Theorem: {Net Work done on object} = {change in kinetic energy of object}

Example A 200g hockey puck initially at rest on ice is pushed by a hockey stick by a constant force of 6.0N. What is the hockey puck’s speed after it has moved 2m? A 200g hockey puck initially at rest on ice is pushed by a hockey stick by a constant force of 6.0N. What is the hockey puck’s speed after it has moved 2m?

Work & Energy Question Two blocks have masses m 1 and m 2, where m 1 > m 2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e.  > 0) which slows them down to a stop. Which one will go farther before stopping? Two blocks have masses m 1 and m 2, where m 1 > m 2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e.  > 0) which slows them down to a stop. Which one will go farther before stopping? (a) (b) (c) (a) m 1 (b) m 2 (c) they will go the same distance m1m1 m2m2

Solution The work-energy theorem says that for any object W NET =  K The work-energy theorem says that for any object W NET =  K In this example the only force that does work is friction (since both N and mg are perpendicular to the block’s motion). In this example the only force that does work is friction (since both N and mg are perpendicular to the block’s motion). m f N mg

Solution The work-energy theorem says that for any object W NET =  K The work-energy theorem says that for any object W NET =  K In this example the only force that does work is friction (since both N and mg are perpendicular to the blocks motion). In this example the only force that does work is friction (since both N and mg are perpendicular to the blocks motion). The net work done to stop the box is - fD = -  mgD. The net work done to stop the box is - fD = -  mgD. m D è This work “removes” the kinetic energy that the box had: è W NET = K 2 - K 1 = 0 - K 1

Solution The net work done to stop a box is - fD = -  mgD. The net work done to stop a box is - fD = -  mgD. This work “removes” the kinetic energy that the box had: This work “removes” the kinetic energy that the box had: W NET = K 2 - K 1 = 0 - K 1 W NET = K 2 - K 1 = 0 - K 1 This is the same for both boxes (same starting kinetic energy). This is the same for both boxes (same starting kinetic energy).  m 2 gD 2  m 1 gD 1 m 2 D 2  m 1 D 1 m1m1 D1D1 m2m2 D2D2 Since m 1 > m 2 we can see that D 2 > D 1

A Simple Application: Work done by gravity on a falling object What is the speed of an object after falling a distance H, assuming it starts at rest? What is the speed of an object after falling a distance H, assuming it starts at rest? W g = F   r = mg  r cos(0) = mgH W g = F   r = mg  r cos(0) = mgH W g = mgH W g = mgH Work/Kinetic Energy Theorem: W g = mgH = 1 / 2 mv 2 rrrr gmggmg H j v 0 = 0 v

POWER Simply put, power is the rate at which work gets done (or energy gets transferred). Suppose you and I each do 1000J of work, but I do the work in 2 minutes while you do it in 1 minute. We both did the same amount of work, but you did it more quickly (you were more powerful) J J s s Watt (W)

Power = change in energy time Measured in Watts (J/s or W) or in horsepower Ex. 1 horsepower = 746 W or 0.75 kW Vehicle horsepower can be calculated using its RPM & Torque (in the manual). Watt determined that the average horse could do 33, 000 foot-pounds of work per minute (Ex. Move 1000 lbs of coal 33 ft every minute). The equation for power is: P = ΔE = W Δt Δt

Power is a measure of: the change in energy over time. aka. Energy used or Work done over time Kahn Academy: Speed of Weight-Lifter:

Power Power is also needed for acceleration and for moving against the force of gravity. The average power can be written in terms of the force and the average velocity:

Understanding A mover pushes a large crate (m= 75 kg) from one side of a truck to the other side ( a distance of 6 m), exerting a steady push of 300 N. If she moves the crate in 20 s, what is the power output during this move?

Understanding What must the power output of an elevator motor be such that it can lift a total mass of 1000 kg, while maintaining a constant speed of 8.0 m/s?