1. Copyright © 2010 Pearson Education, Inc. Lecture Outline Chapter 7 Physics, 4 th Edition James S. Walker

2. Copyright © 2010 Pearson Education, Inc. Chapter 7 Work and Kinetic Energy

3. Copyright © 2010 Pearson Education, Inc. Units of Chapter 7 Work Done by a Constant Force Kinetic Energy and the Work-Energy Theorem Work Done by a Variable Force Power

4. Copyright © 2010 Pearson Education, Inc. 7-1 Work Done by a Constant Force The definition of work, when the force and displacement are in the same direction. (7-1) SI unit: newton-meter (N·m) = joule, J 1 joule = 1 J = 1 N.m = 1 (kg.m/s 2 ).m = 1 kg.m 2 / s 2.

5. Copyright © 2010 Pearson Education, Inc. 7-1 Work Done by a Constant Force

6. Copyright © 2010 Pearson Education, Inc. 7-1 Work Done by a Constant Force If the force is at an angle to the displacement or when the angle between a constant force and the displacement is θ : (7-3)

7. Copyright © 2010 Pearson Education, Inc. 7-1 Work Done by a Constant Force The work can also be written as the dot product of the force and the displacement:

8. Copyright © 2010 Pearson Education, Inc. 7-1 Work Done by a Constant Force EXAMPLE 7-2 - In a gravity escape system (GES), an enclosed lifeboat on a large ship is deployed by letting it slide down a ramp and then continuing in free fall to the water below. Suppose a 4970 kg lifeboat slides a distance of 5.00 m on a ramp, dropping through a vertical height of 2.50 m. How much work does gravity do on the boat? Work done on the lifeboat by gravity is W = Fdcos θ Fcos θ = mg(h/d) = (4970 kg)(9.81 m/s 2 )(2.50 m / 5.0) = 24,400 N W = ( Fcos θ) d = (24,400 N) (5.0 m) = 122,000 J OR W = ( Fcos θ) d = (mg)(d)(h/d) W = mgh = (4970 kg)(9.81 m/s2)(2.50 m) = 122,000 J

9. Copyright © 2010 Pearson Education, Inc. 7-1 Work Done by a Constant Force The work done may be positive, zero, or negative, depending on the angle between the force and the displacement: (i)Work is positive if the force has a component in the direction of motion (-90 o < θ < 90 O ) (ii)Work is zero if the force has no component in the direction of motion (θ = +- 90 O ) (iii)Work is negative if the force has a component opposite to the direction of motion (-90 o < θ < 270 O ) Whenever we calculate work, we must be careful about its sign and not just assume it to be positive. When more than one force acts on an object, the total work is the sum of the work done by each force separately. Thus if force F 1 does work W 1, force F 2 does work W 2, and so on, the total work is W total = W 1 + W 2 +W 3 + ………. = ∑ W Equivalently, the total work can be calculated by first performing a vector sum of all the forces acting on an object to obtain F total and then using our basic definition of work: W tota l = (F total cos θ)d = F total d (cos θ)

10. Copyright © 2010 Pearson Education, Inc. 7-1 Work Done by a Constant Force If there is more than one force acting on an object, we can find the work done by each force, and also the work done by the net force: (7-5)

11. Copyright © 2010 Pearson Education, Inc. 7-1 Work Done by a Constant Force (7-5) A car of mass m coasts down a hill inclined at an angle θ below the horizontal. The car is acted on by three forces: (i) the normal force N exerted by the road (ii) a force due to air resistance, F air and (iii) the force of gravity, mg. Find the total work done in the car as it travels a distance d along the road. W N = Nd cos θ = Nd cos 90 o = Nd (0) = 0 W air = F air d cos 180 o = F air d(-1) = -F air d W total = mgd cos (90 o - θ) = mgd sin θ W total = W N + W air + W mg = 0 – F air d + mgd sin θ

12. Copyright © 2010 Pearson Education, Inc. 7-2 Kinetic Energy and the Work-Energy Theorem When positive work is done on an object, its speed increases; when negative work is done, its speed decreases. Apple falling: W > 0, speed increases Apple tossed upward: W < 0, speed decreases

13. Copyright © 2010 Pearson Education, Inc. 7-2 Kinetic Energy and the Work-Energy Theorem After algebraic manipulations of the equations of motion, we find: Therefore, we define the kinetic energy: (7-6) SI unit : kg.m 2 / s 2 = Joule, J

14. Copyright © 2010 Pearson Education, Inc. 7-2 Kinetic Energy and the Work-Energy Theorem Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy. (7-7)

15. Copyright © 2010 Pearson Education, Inc. 7-2 Kinetic Energy and the Work-Energy Theorem A 4.10 kg box of books is lifted vertically from rest a distance of 1.60 m with a constant, upward applied force of 52.7 N. find (a) the work done by the applied force, (b) the work done by gravity, and ( c) the final speed of the box. Work done by the applied forceW applied force = F app cos 0 o ∆y = (52.7 N)(1.60 m) = 84.3 J Work done by gravity W g = mg cos 180 o ∆y = (4.10 kg)(9.81 m/s 2 )(-1)(1.60 m) = -64.4 J Total Work done on the book W total = W applied force + W g = 84.3 J – 64.4 J = 19.9 J C.The final speed, V f : W total = 1/2mv f 2 - 1/2mv i 2 v f = (2W total /m) 1/2 = (2 (19.9 J) / 4.10 kg) 1/2 V f = 3.12 m/s

16. Copyright © 2010 Pearson Education, Inc. 7-3 Work Done by a Variable Force If the force is constant, we can interpret the work done graphically: Graphical representation of the work done by a constant force A constant force F acting through a distance d does a work W = Fd. Note that Fd is also equal to the shaded area between the force line and the x-axis.

17. Copyright © 2010 Pearson Education, Inc. 7-3 Work Done by a Variable Force If the force takes on several successive constant values: Work done by a non constant force (a) A force with a value F 1 from 0 to x 1 and a value F 2 from x 1 to x 2 does the work W = F 1 x 1 + F 2 (x 2 – x 1 ) this is simply the area of the two shaded rectangles. (b). If a force takes on a number of different values, the work it does is still the total area between the force lines and the x-axis, just as in part (a).

18. Copyright © 2010 Pearson Education, Inc. 7-3 Work Done by a Variable Force We can then approximate a continuously varying force by a succession of constant values.

19. Copyright © 2010 Pearson Education, Inc. 7-3 Work Done by a Variable Force The force needed to stretch a spring an amount x is F = kx. Therefore, the work done in stretching the spring is (7-8)

20. Copyright © 2010 Pearson Education, Inc. 7-4 Power Power is a measure of the rate at which work is done: (7-10) SI unit: J/s = watt, W 1 horsepower = 1 hp = 746 W

21. Copyright © 2010 Pearson Education, Inc. 7-4 Power

22. Copyright © 2010 Pearson Education, Inc. 7-4 Power If an object is moving at a constant speed in the face of friction, gravity, air resistance, and so forth, the power exerted by the driving force can be written: (7-13)

23. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 7 If the force is constant and parallel to the displacement, work is force times distance If the force is not parallel to the displacement, The total work is the work done by the net force:

24. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 7 SI unit of work: the joule, J Total work is equal to the change in kinetic energy: where

25. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 7 Work done by a spring force: Power is the rate at which work is done: SI unit of power: the watt, W