Chapter 6 Work and Energy Work (dot product) Work by gravity, by spring Kinetic energy, power Work-kinetic energy theorem C.M. system 4/22/2017 Phys 201, Spring 2011
Definition of Work: Constant Force Ingredients: Force (F), displacement (Δr) Work, W, of a constant force F acting through a displacement Δr : Vector “Dot Product” Work is a scalar. Units are Newton meter: Joule = N x m 4/22/2017 Phys 201, Spring 2011
Work and Force Direction Work is done in lifting the box (you do +work; gravity -) No work is done on the bucket when held still, or to move horizontally 4/22/2017 Phys 201, Spring 2011
More on “dot product” (or scalar product) θ a b ba Definition: a.b = ab cos θ = a[b cos θ] = aba = b[a cos θ] = 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 !! θ a ab b 4/22/2017 Phys 201, Spring 2011
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 4/22/2017 Phys 201, Spring 2011
More 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 uniform circular motion): 4/22/2017 Phys 201, Spring 2011
Which of the statements below is correct? The scalar product of two vectors can be negative. AcB=c(BA), where c is a constant. The scalar product can be non-zero even if two of the three components of the two vectors are equal to zero. If A = B+C, then DA = DB+DC All of the above statements are correct. 4/22/2017 Phys 201, Spring 2011
What about multiple forces? Suppose FNET = F1 + F2 and the displacement is Δr. The work done by each force is: W1 = F1 . Δr W2 = F2 . Δr WTOT = W1 + W2 = F1 . Δr + F2 . Δr = (F1 + F2 ) . Δr WTOT = FTOT . Δr It’s the total force that matters!! FNET F1 Δr F2 4/22/2017 Phys 201, Spring 2011
Question 1 W T FN V You are towing a car up a hill with constant velocity. The work done on the car by the normal force is: 1. positive 2. negative 3. zero correct Normal force is perpendicular to displacement cos θ = 0 4/22/2017 Phys 201, Spring 2011
Question 2 W T FN V You are towing a car up a hill with constant velocity. The work done on the car by the gravitational force is: 1. positive 2. negative 3. zero correct There is a non-zero component of gravitational force pointing opposite the direction of motion. 4/22/2017 Phys 201, Spring 2011
Question 3 W T FN V You are towing a car up a hill with constant velocity. The work done on the car by the tension force is: 1. positive 2. negative 3. zero correct T is pointing in the direction of motion - therefore, work done by this force is positive. 4/22/2017 Phys 201, Spring 2011
Question 4 W T FN V You are towing a car up a hill with constant velocity. The total work done on the car by all forces is: 1. positive 2. negative 3. zero correct Constant velocity implies that there is no net force acting on the car, so there is no work being done overall 4/22/2017 Phys 201, Spring 2011
Lifting an object: A 3000–kg truck is to be loaded onto a ship by a crane that exerts an upward force of 31 kN on the truck. This force, which is strong enough to overcome the gravitational force and keep the truck moving upward, is applied over a distance of 2.0 m. Find (a) the work done on the truck by the crane, (b) the work done on the truck by gravity, (c) the net work done on the truck. (a) the work done by the crane: Wc = Fc d = 31 kN x 2 m = 62 kJ (b) the work done by gravity: Wg = Fg d = (-31 kN) x 2 m = -62 kJ (c) the net work done on the truck: W = Wc + Wg = 0 4/22/2017 Phys 201, Spring 2011
Work by a variable force: Total work, W, of a force F acting through a displacement Δx = dx î is an integration: Fx Fi x Δxi Work = Area under F(x) curve 4/22/2017 Phys 201, Spring 2011
Work done by a spring force Force exerted by a spring extended a distance x is: Fx = -kx (Hooke’s law) Force done by spring on an object when the displacement of spring is changed from xi to xf is: 4/22/2017 Phys 201, Spring 2011
A person pushes against a spring, the opposite end of which is attached to a fixed wall. The spring compresses. Is the work done by the spring on the person positive, negative or zero? positive negative zero 4/22/2017 Phys 201, Spring 2011
It is convertible into other forms without loss (i.e it is conserved) Energy Energy is that quality of a substance or object which “causes something to happen”; or “capability of exerting forces”; or “ability to do work”… The vagueness of the definition is due to the fact that energy can result in many effects. Electrical Energy Chemical Energy Mechanical Energy Nuclear Energy The force that accompanies the “capability” takes various forms (labeling the energy). Energy (capability) can increase or decrease. That is, energy can be decreased by exerting a force that does work, and can be increased if an external force does work. It is convertible into other forms without loss (i.e it is conserved) 4/22/2017 Phys 201, Spring 2011
Kinetic Energy The “energy of motion”. W = mV - m V W = F d K.E. = mV Work done on the object increases its energy, -- by how much? (i.e. how to calculate the value?) W = F d V = V + 2 a d 2 F = ma a = 2 d V - V 2 W = (ma) d Turn off sound 2 d V - V 2 W = m d W = mV - m V 1 2 K.E. = mV 1 2 4/22/2017 Phys 201, Spring 2011
Definition of Kinetic Energy Kinetic energy (K.E.) of a particle of mass m moving with speed v is defined as K.E. = ½mv2 Kinetic energy is a useful concept because of the Work/Kinetic Energy theorem, which relates the work done on an object to the change in kinetic energy. 4/22/2017 Phys 201, Spring 2011
A falling object What is the speed of an object that starts at rest and then falls a vertical distance H? Work done by gravitational force WG = FΔr = mgH Work/Kinetic Energy Theorem: v0 = 0 mg j r H v 4/22/2017 Phys 201, Spring 2011
A skier of mass 50 kg is moving at speed 10 m/s at point P1 down a ski slope with negligible friction. What is the skier’s kinetic energy when she is at point P2, 20 m below P1? 2500 J 9800 J 12300 J 13100 J 15000 J Work-kinetic energy theorem: KEf – KEi = Wg = mgH KEf = ½×50×102 + 50×9.8×20 = 2500 J + 9800 J 4/22/2017 Phys 201, Spring 2011
Energy and Newton’s Laws The importance of mechanical energy in classical mechanics is a consequence of Newton’s laws. The concept of energy turns out to be even more general than Newton’s Laws. 4/22/2017 Phys 201, Spring 2011
Problem: Work and 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 with μ>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 Solution next page m1 m2 4/22/2017 Phys 201, Spring 2011
Problem: Work and Energy (Solution) The net work done to stop a box is -fD = -μmgD. The work-kinetic energy theorem says that for any object, WNET = ΔK so WNET=Kf-Ki=0-Ki. Since the boxes start out with the same kinetic energy, we have μm1gD1=μm2gD2 and D1/D2=m2/m1. Since m1>m2, we must have D2>D1. m2 D2 m1 D1 4/22/2017 Phys 201, Spring 2011
Problem: The magnitude of the single force acting on a particle of mass m is given by F = bx2, where b is a constant. The particle starts from rest at x=0. After it travels a distance L, determine its (a) kinetic energy and (b) speed. work done by force: So (a) kinetic energy is K=bL3/3 (b) Since ½mv2=bL3/3, 4/22/2017 Phys 201, Spring 2011
Power: P: Work done by unit time. Units: Watt 4/22/2017 Phys 201, Spring 2011
Question A sports car accelerates from zero to 30 mph in 1.5 s. How long does it take for it to accelerate from zero to 60 mph, assuming the power of the engine to be independent of velocity and neglecting friction? Power is constant, so , a constant. Integrating with respect to time, and noting that the initial velocity is zero, one gets . So getting to twice the speed takes 4 times as long, and the time to reach 60 mph is 4×1.5 = 6s. 4/22/2017 Phys 201, Spring 2011
If a fighter jet doubles its speed, by what factor should the power from the engine change? (Assume that the drag force on the plane is proportional to the square of the plane’s speed.) by half unchanged doubled quadrupled 8 times Magnitude of power is Fv. When the velocity v is doubled, the drag force goes up by a factor of 4, and Fv goes up by a factor of 8 4/22/2017 Phys 201, Spring 2011
Recall: Center of Mass Definition of the center of mass: Because , 4/22/2017 Phys 201, Spring 2011
Net work done on collection Center-Of-Mass Work For systems of particles that are not all moving at the same velocity, there is a work-kinetic energy relation for the center of mass. So where is the translational kinetic energy Net work done on collection of objects change in translational kinetic energy of system = 4/22/2017 Phys 201, Spring 2011