Conservation of Energy Chapter 8 Conservation of Energy HW5:Chapter7: Pb.16, Pb.19,Pb.46 Chapter 8: Pb.20, Pb.23, Pb.33, Pb.68, Pb.71 Due Wednesday, Oct.31 Chapter Opener. Caption: A polevaulter running toward the high bar has kinetic energy. When he plants the pole and puts his weight on it, his kinetic energy gets transformed: first into elastic potential energy of the bent pole and then into gravitational potential energy as his body rises. As he crosses the bar, the pole is straight and has given up all its elastic potential energy to the athlete’s gravitational potential energy. Nearly all his kinetic energy has disappeared, also becoming gravitational potential energy of his body at the great height of the bar (world record over 6 m), which is exactly what he wants. In these, and all other energy transformations that continually take place in the world, the total energy is always conserved. Indeed, the conservation of energy is one of the greatest laws of physics, and finds applications in a wide range of other fields.

Problem 16 (II) A 72-kg trampoline artist jumps vertically upward from the top of a platform with a speed of 4.5m/s. (a) How fast is he going as he lands on the trampoline, 2.0 m below (b) If the trampoline behaves like a spring of spring constant 5.8 × 104 N/m, how far does he depress it?

Problem 21 21.(II) When a mass m sits at rest on a spring, the spring is compressed by a distance d from its undeformed length (Fig. 8–33a). Suppose instead that the mass is released from rest when it barely touches the undeformed spring (Fig. 8–33b). Find the distance D that the spring is compressed before it is able to stop the mass. Does D=d If not, why not?

Problem 22 22. (II) Two masses are connected by a string as shown in Fig. 8–34. Mass mA=4.0kg rests on a frictionless inclined plane, while mB=5.0kg is initially held at a height of h=0.75m above the floor. (a) If mB is allowed to fall, what will be the resulting acceleration of the masses? (b) If the masses were initially at rest, use the kinematic equations (Eqs. 2–12) to find their velocity just before mB hits the floor. (c) Use conservation of energy to find the velocity of the masses just before mB hits the floor. You should get the same answer as in part (b).

8-5 The Law of Conservation of Energy Non-conservative, or dissipative forces: Friction do not conserve mechanical energy. The energy is transformed in: Heat Electrical energy Chemical energy and more However, when these forces are taken into account, the total energy is still conserved:

8-5 The Law of Conservation of Energy The law of conservation of energy is one of the most important principles in physics. The total energy is neither increased nor decreased in any process. Energy can be transformed from one form to another, and transferred from one object to another, but the total amount remains constant.

8-6 Energy conservation with dissipative Forces The law of conservation of total energy is more powerful than work-energy principle that’s is not a statement of conservation of energy.

8-5 The Law of Conservation of Energy Problem 34 (II) Suppose the roller-coaster car in the Fig. 8–32 passes point 1 with a speed of 1.70m/s. If the average force of friction is equal to 0.23 of its weight, with what speed will it reach point 2? The distance traveled is 45.0 m.

8-7 Gravitational Potential Energy and Escape Velocity Far from the surface of the Earth, the force of gravity is not constant: The work done on an object moving in the Earth’s gravitational field is given by: Figure 8-19. Caption: Arbitrary path of particle of mass m moving from point 1 to point 2.

8-7 Gravitational Potential Energy and Escape Velocity Because the value of the work depends only on the end points, the gravitational force is conservative and we can define gravitational potential energy: Figure 8-20. Caption: Gravitational potential energy plotted as a function of r, the distance from Earth’s center. Valid only for points r > rE , the radius of the Earth.

8-7 Gravitational Potential Energy and Escape Velocity Example 8-12: Package dropped from high-speed rocket. A box of empty film canisters is allowed to fall from a rocket traveling outward from Earth at a speed of 1800 m/s when 1600 km above the Earth’s surface. The package eventually falls to the Earth. Estimate its speed just before impact. Ignore air resistance. Solution: Use conservation of energy; v = 5320 m/s. (Air resistance will reduce this considerably, of course!)

8-7 Gravitational Potential Energy and Escape Velocity Problem 50: (II) Two Earth satellites, A and B, each of mass of 950kg are launched into circular orbits around the Earth’s center. Satellite A orbits at an altitude of 4200 km, and satellite B orbits at an altitude of 12,600 km. (a) What are the potential energies of the two satellites? (b) What are the kinetic energies of the two satellites? (c) How much work would it require to change the orbit of satellite A to match that of satellite B? The radius of the earth is 6380km, and the mass of the earth is 5.98X1024kg

8-7 Gravitational Potential Energy and Escape Velocity If an object’s initial kinetic energy is equal to the potential energy at the Earth’s surface, its total energy will be zero. The velocity at which this is true is called the escape velocity; for Earth: it is also the minimum velocity that prevent an object from returning to earth.

8-8 Power Power is the rate at which work is done. Average power: Instantaneous power: In the SI system, the units of power are watts:  

8-8 Power Power is the rate at which energy is transformed In the British system, the basic unit for power is the foot-pound per second, but more often horsepower is used: 1 hp = 550 ft·lb/s = 746 W. Units: Joules/s or Watts, W Puzzler: kilowatt·hours [kW·h] are units of what quantity?

8-8 Power Example 8-14: Stair-climbing power. A 60-kg jogger runs up a long flight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m. (a) Estimate the jogger’s power output in watts and horsepower. (b) How much energy did this require? Figure 8-21. Solution: a. The work is the change in potential energy, and the power is the work divided by time: P = 660 W = 0.88 hp. b. The energy is 2600 J.

8-8 Power Example 8-15: Power needs of a car. Calculate the power required of a 1400-kg car under the following circumstances: (a) the car climbs a 10° hill (a fairly steep hill) at a steady 80 km/h; and (b) the car accelerates along a level road from 90 to 110 km/h in 6.0 s to pass another car. Assume that the average retarding force on the car is FR = 700 N throughout. Figure 8-22. Caption: Example 8–15: Calculation of power needed for a car to climb a hill. Solution: a. If the speed is constant, the forces on the car must add to zero. This gives a force of 3100 N. Then the power is 6.8 x 104 W = 91 hp. b. The car must exert enough force to reach the given acceleration, as well as overcome the retarding force. This gives F = 2000 N. The maximum power is required at the maximum speed, 6.12 x 104 W or 82 hp.

Chapter 9 : Linear Momentum

9-1 Momentum and Its Relation to Force Momentum is the property of a moving object to continue moving Momentum is a vector symbolized by the symbol and is defined as The rate of change of momentum is equal to the net force: This can be shown using Newton’s second law.

Momentum, p Vector units: kg.m/s Bowling Ball vs. Tennis Ball Mass Speed 9 m/s 60 m/s momentum 63kg.m/s 3.42kg.m/s

9-1 Momentum and Its Relation to Force Example: A system consists of three particles with these masses and velocities: mass 3.0 kg moving west at 5.0 m/s; mass 4.0 kg moving west at 10.0 m/s; and mass 5.0 kg moving east at 20.0 m/s. What is total momentum of the system?

9-1 Momentum and Its Relation to Force Example 9-1: Force of a tennis serve. For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mass of 0.060 kg and is in contact with the racket for about 4 ms (4 x 10-3 s), estimate the average force on the ball. Would this force be large enough to lift a 60-kg person? Figure 9-1. Solution: The average force is the change in momentum divided by the time: 800 N. A 60-kg person weighs about 600 N, so this would be large enough.

9-1 Momentum and Its Relation to Force Example 9-2: Washing a car: momentum change and force. Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car, which stops it. (That is, we ignore any splashing back.) What is the force exerted by the water on the car? Figure 9-2. Solution: Every second, 1.5 kg of water hits the car and stops. The force is the change in momentum (final momentum is zero) divided by the time, or 30 N.

9-2 Conservation of Momentum During a collision, measurements show that the total momentum does not change: Figure 9-3. Caption: Momentum is conserved in a collision of two balls, labeled A and B.

9-2 Conservation of Momentum Conservation of momentum can also be derived from Newton’s laws. A collision takes a short enough time that we can ignore external forces. Since the internal forces are equal and opposite, the total momentum is constant. Figure 9-4. Caption: Collision of two objects. Their momenta before collision are pA and pB, and after collision are pA’ and pB’. At any moment during the collision each exerts a force on the other of equal magnitude but opposite direction.

9-2 Conservation of Momentum This is the law of conservation of linear momentum: when the net external force on a system of objects is zero, the total momentum of the system remains constant. Equivalently, the total momentum of an isolated system remains constant.

9-2 Conservation of Momentum Example 9-4: Rifle recoil. Calculate the recoil velocity of a 5.0-kg rifle that shoots a 0.020-kg bullet at a speed of 620 m/s. Figure 9-7. Solution: Use conservation of momentum. The recoil velocity of the gun is -2.5 m/s.

9-2 Conservation of Momentum Problem 12: (I) A 130-kg tackler moving at 2.5m/s meets head-on (and tackles) an 82-kg halfback moving at 5.0m/s. What will be their mutual speed immediately after the collision?

9-3 Collisions and Impulse During a collision, objects are deformed due to the large forces involved. Since , we can Write Integrating, Figure 9-8. Caption: Tennis racket striking a ball. Both the ball and the racket strings are deformed due to the large force each exerts on the other.

9-3 Collisions and Impulse This quantity is defined as the impulse, J: The impulse is equal to the change in momentum: This equation is true if F is the net impulsive force of the object that is much larger than any other force in a short interval of time.

9-3 Collisions and Impulse Since the time of the collision is often very short, we may be able to use the average force, which would produce the same impulse over the same time interval. Figure 9-9. Caption: Force as a function of time during a typical collision: F can become very large; Δt is typically milliseconds for macroscopic collisions. Figure 9-10. Caption: The average force acting over a very brief time interval gives the same impulse (FavgΔt) as the actual force.

Happy ball and sad ball are each dropped onto the floor from the same height. Happy ball bounces back up, sad ball doesn’t. The impulse is? A) bigger for happy B) bigger for sad C) The same for both Question

9-3 Collisions and Impulse Example 9-6: Karate blow. Estimate the impulse and the average force delivered by a karate blow that breaks a board a few cm thick. Assume the hand moves at roughly 10 m/s when it hits the board. Figure 9-11. Solution: Take the mass of the hand plus a reasonable portion of the arm to be 1 kg; if the speed goes from 10 m/s to zero in 1 cm the time is 2 ms. This gives a force of 5000 N.