SPH4U: Introduction to Work 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

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

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”...

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 & Energy Understanding A box is pulled up a rough (  > 0) incline by a rope-pulley-weight arrangement as shown below. A box is pulled up a rough (  > 0) incline by a rope-pulley-weight arrangement as shown below. How many forces are doing work on the box? How many forces are doing work on the box? (a) (a) 2 (b) (b) 3 (c) (c) 4 Box

Solution Draw FBD of box: N f mg T l Consider direction of motion of the box v l Any force not perpendicular to the motion will do work: N does no work (perp. to v) T does positive work f does negative work mg does negative work 3 forces do work

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

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} l University will prove this for a variable force later.

Nice to Know A B The work in going from A to B.

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

What about multiple forces? FFF Suppose F NET = F 1 + F 2 and the r displacement is  r. The work done by each force is: F  r F  r W 1 = F 1   r W 2 = F 2   r W TOT = W 1 + W 2 F  r F  r = F 1   r + F 2   r FF  r = (F 1 + F 2 )   r F  rtotal W TOT = F TOT   r It’s the total force that matters!! F F NET rrrr FF1FF1 FF2FF2

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

Comments... W = F   r No work done if  = 90 o. No work done if  = 90 o. No work done by T. No work done by T. No work done by N. T v v N

Work & Energy Question An inclined plane is accelerating with constant acceleration a. A box resting on the plane is held in place by static friction. How many forces are doing work on the block? An inclined plane is accelerating with constant acceleration a. A box resting on the plane is held in place by static friction. How many forces are doing work on the block? a (a) (b) (c) 3 (a) 1(b) 2(c) 3

Solution First, draw all the forces in the system: First, draw all the forces in the system: a mg N FSFSFSFS

a mg NSolution Recall that W = F  Δr so only forces that have a component along the direction of the displacement are doing work. Recall that W = F  Δr so only forces that have a component along the direction of the displacement are doing work. FSFSFSFS l The answer is (b) 2.

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 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?

Flash