Physics 111: Lecture 9 Today’s Agenda Work & Energy Discussion Definition Dot Product Work of a constant force Work/kinetic energy theorem Work of multiple constant forces Comments

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

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

Mass = Energy (but not in Physics 111) Particle Physics: E = 1010 eV + 5,000,000,000 V e- - 5,000,000,000 V e+ (a) (b) M E = MC2 ( poof ! ) (c)

Energy Conservation Wilberforce Returning Can Energy cannot be destroyed or created. Just changed from one form to another. We say energy is conserved! True for any isolated 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-breaks-road-atmosphere” system is the same. The energy of the car “alone” is not conserved... It is reduced by the braking. Doing “work” on an isolated system will change its “energy”...

Definition of Work: Ingredients: Force (F), displacement (r) Work, W, of a constant force F acting through a displacement r is: W = F r = F r cos  = Fr r F  r Fr displacement “Dot Product”

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

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

Aside: Examples of dot products y z i j k i . i = j . j = k . k = 1 i . j = j . k = k . i = 0 Suppose Then a = 1 i + 2 j + 3 k b = 4 i - 5 j + 6 k a . b = 1x4 + 2x(-5) + 3x6 = 12 a . a = 1x1 + 2x2 + 3x3 = 14 b . b = 4x4 + (-5)x(-5) + 6x6 = 77

Aside: Properties of dot products Magnitude: a2 = |a|2 = a . a = (ax i + ay j) . (ax i + ay j) = ax 2(i . i) + ay 2(j . j) + 2ax ay (i . j) = ax 2 + ay 2 Pythagorean Theorem!! a ax ay i j

Aside: Properties of dot products Components: a = ax i + ay j + az k = (ax , ay , az) = (a . i, a . j, a . k) Derivatives: Apply to velocity So if v is constant (like for UCM):

Back to the definition of Work: Skateboard Work, W, of a force F acting through a displacement  r is: W = F  r F  r

Lecture 9, Act 1 Work & Energy A box is pulled up a rough (m > 0) incline by a rope-pulley-weight arrangement as shown below. How many forces are doing work on the box? (a) 2 (b) 3 (c) 4

Lecture 9, Act 1 Solution N Draw FBD of box: v T Consider direction of motion of the box v Draw FBD of box: T Any force not perpendicular to the motion will do work: mg does negative work f N does no work (perp. to v) T does positive work 3 forces do work mg f does negative work

Work: 1-D Example (constant force) A force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m. F x Work done by F on box : WF = Fx = F x (since F is parallel to x) WF = (10 N) x (5 m) = 50 Joules (J)

Units: Force x Distance = Work Newton x [M][L] / [T]2 Meter = Joule [L] [M][L]2 / [T]2 N-m (Joule) Dyne-cm (erg) = 10-7 J BTU = 1054 J calorie = 4.184 J foot-lb = 1.356 J eV = 1.6x10-19 J cgs other mks

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 v1 before the push and v2 after the push. v1 v2 F m i x

Work & Kinetic Energy... Since the force F is constant, acceleration a will be constant. We have shown that for constant a: v22 - v12 = 2a(x2-x1) = 2ax. multiply by 1/2m: 1/2mv22 - 1/2mv12 = max But F = ma 1/2mv22 - 1/2mv12 = Fx x F v1 v2 m a i

Work & Kinetic Energy... So we find that 1/2mv22 - 1/2mv12 = Fx = WF Define Kinetic Energy K: K = 1/2mv2 K2 - K1 = WF WF = K (Work/kinetic energy theorem) v2 x F v1 m a i

Work/Kinetic Energy Theorem: {Net Work done on object} = {change in kinetic energy of object} We’ll prove this for a variable force later.

Lecture 9, Act 2 Work & Energy Two blocks have masses m1 and m2, where m1 > m2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e. m > 0) which slows them down to a stop. Which one will go farther before stopping? (a) m1 (b) m2 (c) they will go the same distance m1 m2

Lecture 9, Act 2 Solution The work-energy theorem says that for any object WNET = DK In this example the only force that does work is friction (since both N and mg are perpendicular to the block’s motion). N f m mg

Lecture 9, Act 2 Solution The work-energy theorem says that for any object WNET = DK 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 = -mmgD. This work “removes” the kinetic energy that the box had: WNET = K2 - K1 = 0 - K1 m D

Lecture 9, Act 2 Solution The net work done to stop a box is - fD = -mmgD. This work “removes” the kinetic energy that the box had: WNET = K2 - K1 = 0 - K1 This is the same for both boxes (same starting kinetic energy). mm2gD2 = mm1gD1 m2D2 = m1D1 m1 D1 m2 D2 Since m1 > m2 we can see that D2 > D1

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? Wg = F r = mg r cos(0) = mgH Wg = mgH Work/Kinetic Energy Theorem: Wg = mgH = 1/2mv2 v0 = 0 mg j r H v

What about multiple forces? Suppose FNET = F1 + F2 and the displacement is r. The work done by each force is: W1 = F1 r W2 = F2  r WTOT = W1 + W2 = F1 r + F2 r = (F1 + F2 ) r WTOT = FTOT r It’s the total force that matters!! FNET F1 r F2

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

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

Lecture 9, Act 3 Work & Energy 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) 1 (b) 2 (c) 3

Lecture 9, Act 3 Solution First, draw all the forces in the system: FS mg N FS

Lecture 9, Act 3 Solution Recall that W = F Δr so only forces that have a component along the direction of the displacement are doing work. FS a N mg The answer is (b) 2.

Recap of today’s lecture Work & Energy (Text: 6-1 and 7-4) Discussion Definition (Text: 6-1) Dot Product (Text: 6-2) Work of a constant force (Text: 7-1 and 7-2) Work/kinetic energy theorem (Text: 6-1) Properties (units, time independence, etc.) Work of a multiple forces Comments Look at textbook problems Chapter 6: # 1, 39, 67