Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 6 Part 1 Work and Kinetic Energy

Physics 203 – College Physics I Department of Physics – The Citadel Announcements Read Chapter 6 for next time. You can skip sec Today we will discuss sec. 6.1, 6.3, and 6.10 on Work, Kinetic Energy, and Power. There is a problem set on these sections due Thursday: HW3A Next time, we will discuss sec. 6.4 – 6.9: Potential Energy, Energy Conservation, Non-Conservative Forces. Problem set HW3B is due next Tuesday.

Physics 203 – College Physics I Department of Physics – The Citadel Newton’s Law and Orbits First: an orbit problem… Determine the mass of the sun using the properties of Earth’s orbit. (You can treat it as circular.) Newton’s Gravitational law: F = G M s M e / R 2. Newton’s 2 nd Law: F = M e a = M e v 2 /R Uniform circular motion: v =  R / T G M s M e / R 2 = M e (  /T) 2 R M s = (2  /T) 2 R 3 / G

Physics 203 – College Physics I Department of Physics – The Citadel Newton’s Law and Orbits M s = 4  2 R 3 /(GT 2 ) R = 1.50 × m T = 1 year = 3.16 × 10 7 s G = 6.67 × 10 –11 Nm 2 /kg 2 The numbers give M s = 2.00 × kg

Physics 203 – College Physics I Department of Physics – The Citadel Effect of Force over Distance Applying a force to a particle over distance changes its speed in the direction of the force: v f 2 – v i 2 = 2ax (1 dim) Multiply by ½ m : ½ mv f 2 – ½mv i 2 = max Use F = ma: ½ mv f 2 – ½ mv i 2 = Fx

Physics 203 – College Physics I Department of Physics – The Citadel Definitions: Kinetic Energy = K = ½ mv 2. Work = W = Fx. Units: Work = N m = Joules (J). W ork-energy principle: The work done by the net force on a mass causes a change in kinetic energy:  K = W The Work-Energy Theorem

Physics 203 – College Physics I Department of Physics – The Citadel Work Requires Motion Work is done only when there is motion. W = Fx requires both F and x to be nonzero for W to be nonzero. You can push all day on a wall and get very tired, but if it doesn’t move, you did no work on it.

Physics 203 – College Physics I Department of Physics – The Citadel Question Suppose you apply a force F p = 50 N to a box, which causes it to move at a constant speed through a distance of 10 m. What is the net work done on the box? A) 0.2 J B) 500 J C) 250 J D) 0 J E) 5 J 10 m FfFf F p = 50 N

Physics 203 – College Physics I Department of Physics – The Citadel Example Suppose you apply a force F p = 60 N to a box of mass m = 15 kg initially at rest, with coefficient of kinetic friction  k = 0.3. Two forces act on the box: F p and F f =  k mg. x FpFp FpFp

Physics 203 – College Physics I Department of Physics – The Citadel Example 1. How much work do you do on the box when it moves a distance x = 12 m ? You do an amount of work W p = F p x = (60 N)(12 m) = 720 Nm = 720 J x The pushing force is in the direction of motion, so the work is positive. FpFp

Physics 203 – College Physics I Department of Physics – The Citadel Example 2. How much work does friction do when you move the box 12 m ? The force of friction is F f = –  k mg = – (0.3)(15 kg)(9.8 m/s 2 ) = –44 N The work done by friction is W f = F f x = (–44 N)(12 m) = –528 J x FpFp

Physics 203 – College Physics I Department of Physics – The Citadel Example 3. What is the kinetic energy of the box after being pushed 12 m ? The box was initially at rest, so the kinetic energy is the net work, K = W p + W f = 720 J – 528 J = 192 J 4. What is the speed of the box after being pushed 12 m ? The kinetic energy is K = ½ m v 2. v = √ 2 K/m = √ 2(192 J) / 15 kg = 5.1 m/s

Physics 203 – College Physics I Department of Physics – The Citadel Orbit A satellite of weight mg is in a near-Earth circular orbit. How much work does gravity do on the satellite during each orbit? A) mgR e B)  mgR e C) 0 Think of the work-energy principle:  K = W. ReReReRe

Physics 203 – College Physics I Department of Physics – The Citadel Only the component of force in the direction of motion does work. A force perpendicular to the motion does no work! It can’t change the speed, so it doesn’t affect the kinetic energy. W =  K Work in More than 1 Dimension x y W = (F cos  x F → θ

Physics 203 – College Physics I Department of Physics – The Citadel High Dive via Work/Energy A diver jumps with initial speed v 0 = 1.4 m/s. At what speed does she enter the water 5.0 m below? K f = K 0 + W with K 0 = ½ mv 0 2, K f = ½ mv f 2. Gravity is a constant force in the -y direction, so W = - mg  y = mgh. The x motion does not affect the work. v0v0 vfvf 5.0 m

Physics 203 – College Physics I Department of Physics – The Citadel High Dive via Work/Energy K f = K 0 + W ½ mv f 2 = ½ mv mgh v f = √v gh v 0 = 1.4 m/s, g = 9.8 m/s 2, h = 5.0 m. v f = 10.0 m/s. v0v0 vfvf 5.0 m

Physics 203 – College Physics I Department of Physics – The Citadel Bowling Balls A ball-feeder lifts balls up a 1 m long ramp to a platform 0.5 m above the floor. How much force must the feeder arm exert to lift a ball weighing 50 N ? Enter the answer in Newtons. 1.0 m 0.5 m

Physics 203 – College Physics I Department of Physics – The Citadel Bowling Balls The feeder does work W p = F p L = F × 1.0 m. Gravity does work W g =  mgh =  (50N)(0.5 m) =  25 N W p must be at least 25 J. F p must be at least 25 N. 1.0 m 0.5 m

Physics 203 – College Physics I Department of Physics – The Citadel Power Power is the rate of doing work: P = W/t. If the force F acts in the direction of motion, then P = Fv (instantaneous) These are consistent because x = v t is the distance traveled, so P = F v = F x/t = W/t.

Physics 203 – College Physics I Department of Physics – The Citadel Bowling Balls If the ball feeder is powered by a 5 W motor, how many balls per minute can it lift, in continuous operation? Each ball required W p = 25 J of work. The motor supplies (5 W) (60 s) = 300 J in a minute. That is enough energy to lift 12 balls. 1.0 m 0.5 m

Physics 203 – College Physics I Department of Physics – The Citadel Pulleys How much force do I have to pull with to lift a block of mass M at a constant speed? Solve it using work and energy. FpFp M

Physics 203 – College Physics I Department of Physics – The Citadel Pulleys Pulling a distance L with force F p does work W p = F p L The block moves up a distance ½ L while lifted with a force F Lift = Mg. W p = F p L = W Lift = ½ LMg F p = ½ Mg FpFp F Lift L ½ L M

Physics 203 – College Physics I Department of Physics – The Citadel Work by a Spring When a spring is compressed or stretched, there is a restoring force given by Hooke’s Law, F = – k x. k = spring constant. L x =  L x x = 0 F

Physics 203 – College Physics I Department of Physics – The Citadel Work by a Spring Suppose a ball is placed in front of the spring. If it is held at x = –L and then released, how much work does the spring do on the ball as it returns to its equilibrium position, launching the ball? x =  L x x = 0 m

Physics 203 – College Physics I Department of Physics – The Citadel The work done by a changing force is the average force times the distance. The force decreases from kL down to 0 linearly, so the average force is F = ½ kL. W = F L = ½ kL 2. Work by a Spring W = ½ k L 2 kL 0 F x 0 LL

Physics 203 – College Physics I Department of Physics – The Citadel Spring Gun Suppose a ball of mass m = 250 g is pushed back an distance L = 4 cm on a spring of spring constant k = 120 N/cm and released. How much work is needed to compress the spring? L m L = m k = 12,000 N/m W = ½ kL 2 = 9.60 J.

Physics 203 – College Physics I Department of Physics – The Citadel Spring Gun When the spring is released, what is the launch speed of the ball? Work-Energy Theorem: K = ½ mv 2 = W = 9.60 J. m = kg. v = √2K/m = 8.76 m/s. v m