1. Work, Energy and Power (Chap.6 p.139) Work, Energy and Power A.Prof. Hamid NEBDI Department of Physics Faculty of Applied Science. Umm Al-Qura University جا معة أم القرى Faculty of Applied Science كـلية العلوم التطبيقية مدخل الفيزياء الطبية 104 General Physics 104 (For medical students) مدخل الفيزياء الطبية 104

2. 1.Work Suppose that an object is displaced a distance s, and that a force F acting on the object has a constant component F s along s. Figure1 Figure 2 Figure1 Figure 2 Then the work W done by the force F is defined as the product of the force component F s and the displacement : W = F s. s (1) If F is at an angle Ѳ to s, then F s = F cos Ѳ, and the work can be written as W = F. s. cos Ѳ(2) The S.I unit of work is the joule (J). 1 joule = 1 N.m

3. Example 6.1: A 600 N force is applied by a man to a dresser that moves 2 m. Find the work done if the force and displacement are (a)parallel; (b)at right angles; (c)oppositely directed (Fig.3); we may imagine that the dresser is being slowed and brought to rest. Figure 3

4. Solution 6.1: (a)F and s are parallel, so cos Ѳ = cos 0⁰ = 1 and Eq. 2 gives: W = F. s. cos Ѳ = (600 N) (2m) (1) = 1200 J The man does 1200 J of work on the dresser. Since F is parallel to s, F s =F, and we obtain the same result using Eq. 1. (b) F and s are perpendicular, so cos Ѳ = cos 90⁰ = 0 and Eq. 2 gives: W = F. s. cos Ѳ = (600 N) (2m) (0) = 0 J No work is done when the force is at right angles to the displacement, since F s = 0. (c) F and s are opposite, so cos Ѳ = cos 180⁰ = -1 and Eq. 2 gives: W = F. s. cos Ѳ = (600 N) (2m) (-1) = -1200 J In this case the work done by the force is negative, so the object is doing work on the man. Note that here F is opposite to s, so F s = -F.

5. Example 6.2: A horse pulls a barge along a canal with a rope in which the tension is 1000 N (Fig.4). The rope is at an angle of 10⁰ with the towpath and the direction of the barge. (a)How much work is done by the horse in pulling the barge 100 m upstream at a constant velocity? (b)What is the net force on the barge? Figure 4

6. Solution 6.2: (a)The work done by the constant force T in moving the barge a distance s is given by : W = T. s. cos Ѳ = (1000 N) (100m) (cos 10⁰) = (b)Since the barge moves at a constant velocity, the sum of all forces on it must be zero. There must be another force acting that is not shown in Fig. 4. Figure 4

7. Remark: Remark: - If the force varies in magnitude or direction with respect to the displacement, the correct procedure is to consider the work done in a series of many small displacements. In each piece, we compute : ∆W =, where is the average force in this part of the motion. The sum of all these small terms then gives the net work done. - For computing this work, we can use the graphical method (Fig. 5); the area under the graph of F s versus s for any motion is the work done. Figure 5

8. 2. Kinetic Energy - The kinetic energy of an object is a measure of the work an object can do by virtue of its motion. - The translational kinetic energy K of an object of mass m and velocity v is: (3) - The work done on an object and its kinetic energy obey the following fundamental principle (work-energy principal): The final kinetic energy (K) of an object is equal to its initial kinetic energy (K 0 ) plus the total work (W) done on it by all the forces acting. K = K 0 + W (4) (5) Figure 6

9. Example 6.3: Figure 7a

10. Solution 6.3:

11. 3. Potential Energy - In general, potential energy is energy associated with the position or configuration of a mechanical system. In Fig. 8, a ball rises from initial height h 0 to a height h. The gravitational force mg is opposite in direction to the displacement s = h – h 0, so the work done is negative:W(grav)= - mg (h-h 0 ) The change in potential energy is : ∆ U = U – U 0 (6) The magnitude of ∆ U is defined to be equal to the magnitude of W(grav): U – U 0 = - W(grav) (7) This result for potential energy change involves a difference of two terms on each side, and suggests that we define the potential energy themselves at h and h 0, respectively, by U = mgh and U 0 = mgh 0 (8) Figure 8

12. 4. Total Energy - By the work-energy principal, K = K 0 + W(grav) = K 0 - (U – U 0 ) so we have the important result: K + U = K 0 + U 0 (W a = 0) (9) The notation W a = 0 reminds us that the work done by applied forces is zero; only the gravitational forces is doing work here. - The sum of kinetic energy and the potential energy is called the total mechanical energy:E = K + U(10) Thus Eq. 9 means that when there is no work done by applied forces, the total mechanical energy is constant or conserved. - If applied forces also do a work, Eq. 9 must be generalized to include this work, W a. Then we have: K + U = K 0 + U 0 + W a (11) or E = E 0 + W a (12) The final mechanical energy E = K + U is equal to the initial mechanical energy E 0 = K 0 + U 0 plus the work done by the applied forces.

13. Example 6.5: A woman skis from rest down a hill 20 m height (Fig. 9a). If friction is negligible, what is her speed at the bottom of the slope? Figure 9a Figure 9a

14. Solution 6.5: Figure 9

15. -The work done by conservative forces can be handled conveniently by the introduction of the potential energy concept. - This is not true for frictional forces. There are treated as applied forces: they give negative work. - The work of these forces are converted in thermal energy: dissipative forces. 5. Dissipative Forces

16. A woman skies from rest down a hill 20m high. If friction is not negligible and if her speed at the bottom of the slope is only 1ms -1, how much work is done by frictional forces if her mass is 50kg. Solution 6.6: Example 6.6: Potential energy at the starts from rest: Again, we choose the reference level : bottom of the slope so: Using: E = E 0 + W a, U=0 Kinetic energy at the top is: K 0 =0 U 0 =mgd Kinetic energy at the bottom of the hill is: Thus: We can write: K+U=K 0 +U 0 +Wa Hence:

17. 6. Conservative Forces - Any force that has the property that work it does is the same for all paths between two given points is said to be a conservative force. - This property makes it meaningful to associate a potential energy with a position. - Gravitational, electrical and spring forces are conservative forces. - Friction and many other forces are not conservative. - In Figure 10, the work done by gravity is the same for the paths ABC and AC. Whenever the work done by a force is the same for all paths, the force is said to be conservative, and its effects can be included in the potential energy. Figure 10

18. 7. Power - When an amount of work ∆W is done in a time ∆t, the average power is defined as the average rate of doing work: (10) - The instantaneous power P is found by considering smaller and smaller time intervals, so : (11) - The S.I. power unit is a joule per second, which is called a watt (W). - Energy is often sold by electrical utilities by the kilowatt hour (kW h). This is kilowatt of power for 1 hour. In terms of S.I. units, - Another expression for the power is often useful. The work done by a force F acting through a small displacement ∆s in a short time ∆t is ∆W= F s ∆s. Dividing by ∆t gives the power: -Since ∆s /∆t is the velocity, the power is also given by:

19. Example 6.14: (see p. 152) A 70-kg man runs up a flight of stairs 3m high in 2s. (a) How much work does he do against gravitational forces? (b) what is his average power output? Solution 6.14: