5.4 Exercises 135 Fig 5.45 See Exercise (37) FD t mg y (39) A canonical pendulum consists of a bob of mass m attached to the end of a cord of length The bob whirls around in a horizontal circle of radius r at a constant speed v while the cord always makes an angle θ with the vertical, see Fig 5.46 Show that the bob’s speed v and period T (the time for one complete revolution) are given by: v = rg tan θ = T = 2π g sin θ tan θ , cos θ g Fig 5.46 See Exercise (39) θ θ r m (40) What condition must be imposed on the relationship that governs the period of a canonical pendulum in order to reach to the period of a simple pendulum? Work, Energy, and Power Work, energy, and power are words that have different meanings in our everyday life Nevertheless, physicists give them specific definitions, which we present in this chapter The work-energy power approach provides identical results to those obtained by Newtonian mechanics, but usually with simpler analysis, especially when dealing with complex situations where forces are not constant Therefore, we will introduce two extremely important concepts: the work-energy-theorem and conservation of energy 6.1 Work Done by a Constant Force → Consider a body that experiences a constant force F while undergoing a displacement → s as it moves, see Fig 6.1 We then define the work done by the constant force as follows: Work done by a constant force: Is defined as the product of the component of the force in the direction of the displacement and the magnitude of the displacement Thus: → → W = (F cos θ ) s = F • s = ⎧ ⎪ ⎪ ⎨ +Fs ⎪ ⎪ ⎩ −Fs if θ = 0◦ if θ = 90◦ if θ = (6.1) 180◦ H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_6, © Springer-Verlag Berlin Heidelberg 2013 137 138 Work, Energy, and Power Fig 6.1 The work done by a F → constant force F while undergoing a displacement → s Motion F cos is W = F s cos θ s The unit of work in SI units is N.m [abbreviated by joule (J)], i.e., J = N.m, and in cgs units is dyne.cm (abbreviated by erg), i.e erg = dyne.cm Note that J = 107 erg, see Table 6.1 Table 6.1 Units of work System Unit of work Name of combined unit SI N.m joule (J) cgs dyne.cm erg British ft.lb ft.lb Work Done by a Weight Consider a block of mass m to be lifted up with almost zero acceleration (i.e., a 0) → by a constant force F applied by a person, see Fig 6.2 While in motion, the force → → F and the weight m→ g will be oppositely directed but equal in magnitude, i.e F = −m→ g That is: F = mg Fig 6.2 Lifting a block with (6.2) F almost zero acceleration mg s Lifting Final Initial If the upward displacement of the block is denoted by → s , as in Fig 6.2, then we can → calculate the work done by F as follows: 6.1 Work Done by a Constant Force 139 → WF = F • → s = Fs cos 0◦ = Fs = m g s (6.3) → where we have used the fact that the angle between the two parallel vectors F and → s is zero Also, we can calculate the work done by the gravitational force m → g as follows: Wg = m → g •→ s = m g s cos 180◦ = −m g s (6.4) Thus, we conclude that: WF = m g s and Wg = −m g s (Lifting case) (6.5) where we have used the fact that the angle between the two antiparallel vectors m → g and → s is 180◦ The net work WF +Wg done on the block is zero, as expected, because the net force on the block is zero This is not, of course, to say that it takes no work to lift a block through a vertical height s In such a context, we not refer to the net work, but to the work done by the person When we lower the block vertically downward with almost zero acceleration for → a displacement → s , see Fig 6.3, the sign of the work done by F and m → g will be reversed, since the sign of → s has reversed Following similar steps, one can easily find: → WF = F • → s = F s cos 180◦ = −F s = −m g s (6.6) Wg = m → g •→ s = m g s cos 0◦ = m g s (6.7) Thus, we conclude that: Fig 6.3 Lowering down a and Wg = m g s (Lowering case) (6.8) Initial block with almost zero acceleration F Final mg s Lowering WF = −m g s 140 Work, Energy, and Power Work Done by Friction A common example in which the work is always negative is the work done by → friction When a block slides over a rough surface due to an applied force F , as → shown in Fig 6.4, the work done by the frictional force f k while the block undergoes a displacement → s is: → Wf = f k • → s = fk s cos 180◦ (6.9) = −fk s N F Motion θ Initial fk Final s mg → Fig 6.4 The work done by the kinetic frictional force f k while the block undergoes a displacement → s is always negative and equals Wf = −fk s From Fig 6.4, one can easily find the work done by gravity, the normal force, and the applied force as follows: Wg = m → g •→ s = m g s cos 90◦ = → WN = N • → s = N s cos 90◦ = → WF = F • → s = F s cos θ (6.10) (6.11) (6.12) Example 6.1 A block of mass m is pushed up a rough inclined plane of angle θ by a constant → force F parallel to the incline, as shown in Fig 6.5 The displacement of the block → → up the incline is d (a) Find the work done by: the force F, the kinetic friction → → f k , the force of gravity m → g , and the normal force N (b) Calculate the work done of part (a) for m = kg, μk = 0.5, θ = 30◦ , F = 20 N, and d = m 6.1 Work Done by a Constant Force 141 N d F h mg sθ yf mg co yi sin fk θ θ mg θ Fig 6.5 → → Solution: (a) Since F is in the same direction as the displacement d , we get: → → WF = F • d = F d cos 0◦ = F d The work done by gravity is: → Wg = m → g • d = m g d cos(90◦ + θ ) = −m g d sin θ = −m g h where h = yf − yi = d sin θ is the value of the vertical height That is, the work done by gravity is negative and has a magnitude m g multiplied by height h This result and Eq 6.4 proves that the work is independent of the path taken between any two points → → Since the force of friction f k is opposite to the displacement d , fk = μk N, and N = mg cos θ , the work done by friction will be: → → Wf = f k • d = −fk d = −μk m g d cos θ → → Since N is perpendicular to d ,we get: → → WN = N • d = N d cos 90◦ = (b) Using the values given, the work done by each force will be: WF = F d = (20 N)(5 m) = 100 J Wg = −m g d sin θ = −(2 kg)(9.8 m/s2 )(5 m)(sin 30◦ ) = −49 J Wf = −μk m g d cos θ = −0.5 × (2 kg)(9.8 m/s2 )(5 m)(cos 30◦ ) = −42.4 J Thus: Wnet = WF + Wg + Wf + WN = 100 − 49 − 42.4 + = 8.6 J 142 Work, Energy, and Power 6.2 Work Done by a Variable Force One-Dimensional Analysis Consider an object that is being displaced along the x-axis from xi to xf due to the application of a varying positive force F(x), as shown in Fig 6.6a To calculate the work done by this force, we imagine that the object undergoes a very small displacement x from x to x + x due to the effect of an approximate constant force F(x) as shown in Fig 6.6b For this very small displacement, we represent the amount of work done by the force by the expression: W = F(x) x (6.13) which is just the area of the magnified rectangle shown in Fig 6.6b Then, the total work done from xi to xf by the variable force F(x) is approximately equal to the sum of the large number of rectangles in Fig 6.6b, i.e the total area under the force curve Thus: xf W F(x) x (6.14) xi In the limit where x approaches zero, the value of the sum in the last equation approaches the exact value of the area under the force curve, see Fig 6.6c As you probably know from calculus, the limit of that sum is called an integral and is represented by: xf xf lim x→0 F(x) x= xi F(x) dx F (x) F (x) F (x) F (x) xi xf x xi (6.15) xi xf x (a) x (b) x Area= F (x) x x Work=Area xi xf x (c) Fig 6.6 (a) A variable force F(x) displaces a body in the positive x direction from xi to xf (b) The area under the curve is divided into narrow strips of thickness x, so that the approximate work done by the W = F(x) x (c) In the limiting case, the work done by force F(x) for the small displacement x is the force is the colored area under the force curve 6.2 Work Done by a Variable Force 143 Therefore, we can express the work done by a variable force F(x) on an object that undergoes a displacement from xi to xf as follows: xf W = F(x) dx (6.16) xi If F(x) is positive in some regions and negative in others, the last sum is called the net signed area and is equal to the area of the regions where F(x) > minus the area of the regions where F(x) > Example 6.2 A force acting on an object varies with x as shown in Fig 6.7 Find the work done by the force when the object undergoes a displacement from x = to x = m Fig 6.7 b F(x) (N) H c -2 x h a -4 Solution: The work done by the force equals the net signed area between the curve and the displacement from x = to x = m That is, the area of the trapezoid minus the area of the triangle Thus: W = Area of the trapeziod − Area of the triangle = = 2 (a + b) H − ch × (4 m + m) × (6 N) − = 18 J − J = 12 J × (3 m)(4 N) 144 Work, Energy, and Power Work Done by a Spring A spring is one type of common physical system in which the force (known as the spring force) varies with position Figure 6.8a, shows a massless block on a horizontal frictionless surface attached to the free end of a relaxed spring If the spring is stretched or compressed a small distance from equilibrium, the spring will exert a force on the block This force is given by Hooke’s law as follows: F = − kH x (Hooke’s law) (6.17) where x is the displacement of the block from its equilibrium position (x = 0) and kH is a positive constant known as the spring constant (or the force constant) The negative sign in Hooke’s law indicates that the direction of the force is always opposite to the displacement The spring force is positive (to the right) when x < 0, as in Fig 6.8b, and is negative (to the left) when x > 0, as in Fig 6.8c This type of force always acts toward the equilibrium and is called a restoring force Fig 6.8 The variation of the Equilibrium position F=0 x=0 force of a spring on a block (a) When x = 0, the force is (a) Frictionless zero (equilibrium position) (b) When x is negative, the x x=0 force is positive (compressed F spring) (c) When x is positive, F x the force is negative (stretched spring) (d) Graph of F versus (b) x x The work done by the spring x 21 force as the block moves from −xm to is the colored F triangular area which equals 2 kH xm x=0 F x (c) x x=0 Area H H x 2m x21 F F xm Area (d) xm 38.97 - xm H x x 6.2 Work Done by a Variable Force 145 If we allow the block to compress the spring a distance of xm from its equilibrium position and then release the block, it will move from −xm through the equilibrium position x = to xm In the absence of friction, the block will oscillate indefinitely between −xm and xm In this case xm is called the amplitude of the oscillations To calculate the work done by the spring force on the body as it moves from xi = −xm to xf = 0, we use Hooke’s law in Eq 6.16 as follows: xf Ws = F(x) dx = −xm xi (−kH x) dx = − 21 kH x −xm = 21 kH xm (6.18) Note that the work done by the spring force is positive because the spring force is in the same direction as the displacement We can reach the same result of Eq 6.18 if we plot F versus x, as shown in Fig 6.8d, and then calculate the area of the colored triangle that has a base xm and height kH xm On the other hand, when xi = In this part of the motion, the spring and xf = xm , we can find that Ws = − 21 kH xm force is to the left and the displacement is to the right, resulting in a negative work Generally, if the block undergoes an arbitrary displacement from xi to xf , the work done by the spring force will be given by: xf Ws = xf F(x) dx = xi (−kH x) dx = 21 kH xi2 − 21 kH xf2 (6.19) xi This shows that the work done is zero for any motion that has xi = xf → Let us calculate the work done by the applied force F app when the block moves → very slowly from xi to xf , see Fig 6.9a and b To find this work we notice that F app → is equal and opposite to the spring force F at any displacement, i.e Fapp = −F = −(−kH x) = kH x Thus: xf WFapp = xf Fapp dx = xi kH x dx = xi x f kH x x i = 21 kH xf2 − 21 kH xi2 (6.20) Comparing Eq 6.19 and Eq 6.20 we find that WFapp = − Ws , as expected If we plot Fapp versus x, as shown in Fig 6.9c, then the work done by F in compressing the spring very slowly from xi = to xf = −xm equals the area of the colored triangle (F that has a base xm and height kH xm , i.e Wapp = 21 kH xm app and the displacement are negative) 6.3 Work-Energy Theorem 151 Thus, the acceleration of the box will be given by: F cos θ − μk (mg − F sin θ ) m (50 N)(cos 60◦ ) − 0.2 × [(10 kg)(9.8 m/s2 ) − (50 N)(0.866)] = 1.406 m/s2 = 10 kg a= To find the final speed, we use the kinematic equation vf2 = vi2 + 2as when vi = to get: vf = √ 2as = × (1.406 m/s2 )(4 m) = 3.35 m/s Because the forces are constants in this example, the analysis used by Newtonian mechanics is easier than that of the work-energy theorem 6.4 Conservative Forces and Potential Energy In the previous section we introduced the concept of kinetic energy and found that it can change only if work is done on the object In this section we introduce another form of energy, called potential energy, associated with the position or configuration of an object, and can be thought of as a stored energy that can be converted to kinetic energy or to work We begin by defining the following: (a) Conservative and Non-conservative Forces Conservative Forces In Example 6.1 we were able to see that the work done by gravity depends only on the initial and final vertical coordinates and hence is independent of the path taken between any two points Also, we found the same holds true in the case of a spring In addition, we can easily see from Sect 6.2 that the net work done on the object by the gravitational force during a round trip is zero When a force exhibits these properties, it is called a conservative force With reference to the arbitrary paths of Fig 6.11a, we can write the first condition for a conservative force as: Wab (path 1) = Wab (path 2) (6.34) 152 Work, Energy, and Power b b 1 2 a a (a) (b) Fig 6.11 (a) A conservative force acts on a particle moving from point a to point b by following either path or path (b) A conservative force acts on a particle moving in a round trip from point a to point b along path and then back to point a along path i.e., the work done by a conservative force on a particle moving from a to b along path is the same as from a to b along path In words: Spotlight The net work done by a conservative force on a particle moving between any two points does not depend on the path taken Also, with reference to the arbitrary paths of Fig 6.11b, we can write the second condition for a conservative force as: ⎫ Wab (path 1) = −Wba (path 2) ⎪ ⎪ ⎬ (6.35) or ⎪ ⎪ W (path 1) + W (path 2) = ⎭ ab ba That is, the work done by a conservative force on a particle that moves in a round trip from a to b along path and then from b to a along path is zero In other words: Soptlight The net work done by a conservative force on a particle that is moving around any closed path is zero From the work-energy theorem, W = for a round trip, which means that the particle will return to its starting point with the same kinetic energy it had when it started its motion 6.4 Conservative Forces and Potential Energy 153 We recall from Example 6.1 that the work done by the gravitational force as a particle of mass m moves between two points of elevations yi and yf can be written as: Wg = −mgh = −mg(yf − yi ) (6.36) which satisfies the two conditions of a conservative force Non-conservative Forces Not all forces are conservative For example, let us allow a book to slide across a table that is not frictionless, see Fig 6.12a During the sliding, the kinetic frictional force does negative work on the book, slowing it by transferring energy from its kinetic energy to thermal energy of the book-table system This energy transfer cannot be reversed So, this force is not conservative Therefore, all types of frictional forces are non-conservative forces That is: Soptlight The work done by a non-conservative force on a particle that is moving between any two points depends on the path taken by the particle With reference to the arbitrary paths of Fig 6.12a, we can write the first condition for a non-conservative force as: WAB (path 1) = WAB (path 2) (Non-conservative forces) (6.37) i.e., the work done by a non-conservative force on a particle moving from A to B along path is always not the same along path Also, with reference to the arbitrary paths of Fig 6.12b, we can write the second condition for a non-conservative force as: WAB (path 1) = −WBA (path 2) or ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ WAB (path 1) + WBA (path 2) = ⎭ (6.38) That is, the work done by a non-conservative force on a particle that moves in a round trip from A to B along path and then from B to A along path is not zero 154 Work, Energy, and Power 1 book book B B A A (a) (b) Fig 6.12 (a) The work done by the force of friction depends on the path taken as the book is moved from A to B (b) The work done by the force of friction in a round trip from point A to point B along path and then back to point A along path is not zero (b) Potential Energy We found that the work done by a conservative force is a function of the particle’s initial and final coordinates and neither depends on the path taken nor depends on its velocity Therefore, we can define a function U called the “potential energy” such that the work done by a conservative force equals the decrease of potential energy That is: Wc = − U = Ui − Uf (6.39) where the subscript “c” refers to a conservative force and the change in potential energy is defined as U = Uf − Ui For a particle moving along the x axis under the → effect of a conservative force F that has an x component Fx , we can express Eq 6.39 as follows: f U = Uf − Ui = − Fx dx i or Fx = − dU dx (6.40) It is often convenient to choose some selected initial configuration that has a potential Ui as a reference point and measure all potential energy differences with respect to this point Usually, we set Ui = at this point because it does not really matter what value we assign to Ui Gravitational Potential Energy Consider a particle with mass m moving vertically along the y axis from point yi to point yf Of course, the displacement will be an upward vector while the weight 6.4 Conservative Forces and Potential Energy 155 m→ g will be a downward vector To find the corresponding change in gravitational potential energy of the particle-Earth system, we change the integration in Eq 6.40 to be along the y axis and substitute −mg for the force Fx Thus: yf U = Uf − Ui = − yf (−mg) dy = mg yi That is: dy = mg y yf yi = mg(yf − yi ) yi U = Uf − Ui = mg(yf − yi ) = mg y Only the change in gravitational potential energy (6.41) U is physically important So, according to the previous result, we can set Ui = when yi = This gives: Uf − = mg(yf − 0) which is generally written as follows: U = mgy (Gravitational potential energy) (6.42) That is, the gravitational potential energy associated with the particle-Earth system depends on the vertical position y (or the height) of the particle relative to the reference position y = 0, and does not depend on the horizontal position We can think of U = mgy as the configuration energy stored in the particle-Earth system Elastic Potential Energy Now Consider a block attached to a spring with a spring constant kH as in Fig 6.8 As the block moves from position xi to position xf the spring force F = −kH x does work on the block To find the corresponding change in elastic potential energy of the block-spring system, we substitute −kH x for the force Fx in Eq 6.40 to get: xf xf U = Uf − Ui = − (−kH x) dx = kH xi That is: x dx = kH xi x f 2x x i U = Uf − Ui = 21 kH xf2 − 21 kH xi2 = 21 kH xf2 − 21 kH xi2 (6.43) We set Ui = at the equilibrium position of the block, i.e when xi = This gives: Uf − = 21 kH xf2 − 156 Work, Energy, and Power which is generally written as follows: U = 21 kH x (Elastic potential energy) (6.44) Example 6.5 A ball of mass m = 0.2 kg is at the level of a second balcony which is 10 m above the ground, see Fig 6.13 (a) What is the gravitational potential energy of the ball if we take the reference point y = to be: (1) at the ground, (2) at the first balcony, (3) at the second balcony, and (4) at the top of the building? (b) If the ball drops to the ground, for each of the reference points of part (a), what is the change of potential energy of the ball due to the fall? Fig 6.13 Example 6.5 y y y y 10 10 -5 -5 -10 -10 -15 15 m 15 (1) -5 (2) (3) (4) Solution: (a) Using Eq 6.42, we can calculate the potential energy U of the ball for each choice of y = as follows: coordinate choice (1): coordinate choice (2): U = mgy = (0.2 kg)(9.8 m/s2 )(10 m) = 19.6 J U = mgy = (0.2 kg)(9.8 m/s2 )(5 m) = 9.8 J coordinate choice (3): coordinate choice (4): U = mgy = (0.2 kg)(9.8 m/s2 )(0 m) = J U = mgy = (0.2 kg)(9.8 m/s2 )(−5 m) = − 9.8 J (b) For all the coordinate choices, we have y = −10 m So Eq 6.41 will give the same change in potential energy as follows: 6.4 Conservative Forces and Potential Energy 157 U = mg y = (0.2 kg)(9.8 m/s2 )(−10 m) = −19.6 J Thus, although the value of U depends on the choice of where we let y = 0, the change in potential energy does not In fact, only the change of U, in potential energy is physically important 6.5 U, not the value Conservation of Mechanical Energy When a conservative force does work Wc on a particle, the work-energy theorem tells us that there will be a change in its kinetic energy given by Eq 6.32, which can be rewritten as: Wc = K (6.45) and a change in potential energy given by Eq 6.39, rewritten: Wc = − U (6.46) By equating the last two equations we get: K =− U (6.47) or: K+ U= (K + U) = (6.48) If we define the total mechanical energy E as the sum of the kinetic energy K and potential energy U, i.e E =K +U , (6.49) E=0 (6.50) then Eq 6.48 gives: which is called the principle of conservation of mechanical energy 158 Work, Energy, and Power Conservation of Mechanical Energy: When only a conservative force acts on a system, the kinetic energy and the potential energy can change However, their sum, the mechanical energy E of the system, does not change That is: Ei = Ef (6.51) Ki + Ui = Kf + Uf (6.52) or: If more than one conservative force acts on the system, where each one is associated with a potential energy, then the conservation of mechanical energy will take the form: Ki + Ui = Kf + Uf (6.53) Example 6.6 A frictionless roller-coaster is given a maximum possible initial speed v◦ = m/s when it is at height y◦ = m above the ground and moves freely afterwards, see Fig 6.14 and take g = 10 m/s2 (a) What will be the roller-coaster’s speed when it reaches the lowest point at y1 = m? (b) What will be its maximum height y2 ? o mg yo y2 y1 Fig 6.14 6.5 Conservation of Mechanical Energy 159 Solution: (a) The only force that contributes to the work is the force of gravity Therefore, we can use the law of conservation of mechanical energy Initially, we have Ki = 21 m v◦2 and Ui = m g y◦ Finally, at the lowest point we have Kf = 21 m v12 and Uf = m g y1 Thus, according to Eqs 6.51 and 6.52, we get: Ei = Ef ⇒ Ki + Ui = Kf + Uf ⇒ 1 m v◦2 + m g y◦ = m v12 + m g y1 2 v12 = v◦2 + 2g(y◦ − y1 ) i.e Then: v1 = (6 m/s)2 + 2(10 m/s2 )(6 m − m) = 8.7 m/s (b) The roller-coaster will stop momentarily when it reaches the maximum height y2 , i.e v2 = Accordingly, Eq 6.52 gives: Ei = Ef Then: 6.6 ⇒ Ki + Ui = Kf + Uf y2 = y◦ + 21 v◦2 /g ⇒ ⇒ 2 m v◦ + m g y◦ = 21 m v22 + m g y2 y2 = m + 21 (6 m/s)2 /(10 m/s2 ) = 7.8 m Work Done by Non-conservative Forces In real-life systems, the total mechanical energy is not constant due to the presence of non-conservative forces, such as friction or any applied forces When the work done by all the non-conservative forces on a particle is Wnc and the work done by the conservative force is Wc , then the work-energy theorem tells us that there will be a change in the particle’s kinetic energy given by Eq 6.32 as: Wnc + Wc = K (6.54) Since Eq 6.39 gives Wc = − U, then this equation becomes: Wnc = K+ U = (Kf − Ki ) + (Uf − Ui ) (6.55) Since E = K + U as we saw in Eq 6.49, this equation becomes: Wnc = which generally can be stated as: E = Ef − Ei (6.56) 160 Work, Energy, and Power Spotlight The work done by all non-conservative forces W (or Wnc ) equals the change in the total mechanical energy of the system When there are no non-conservative forces present, Wnc = and hence Ef = Ei ; that is, the total mechanical energy is conserved Example 6.7 A block of initial speed v◦ slides across a floor, see Fig 6.15 A kinetic frictional force of magnitude fk = 50 N does work on the block, stopping it over a displacement of magnitude d = m Find the dissipated mechanical energy Fig 6.15 o Motion fk Final Initial d Solution: From Eq 6.9, the work done by friction is given by: Wnc ≡ Wf = −fk d = −(50 N)(2 m) = −100 J From Eq 6.56 and the above result, the dissipated mechanical energy is: E = Wnc = −100 J Example 6.8 A boy of mass m = 30 kg slides down a curved track of height h = m, see Fig 6.16 The boy starts at point i with a speed vi = and reaches the bottom of the track at point f with a speed vf (a) If the track is frictionless, i.e fk = 0, then find the speed vf (b) If the track is rough and vf = m/s, then find the work done by friction → Solution: (a) The normal force N does no work on the boy since it is always perpendicular to each displacement element on the curved track The only force 6.6 Work Done by Non-conservative Forces 161 that has a change in potential energy is m→ g Therefore, we can use the law of conservation of mechanical energy Initially, we have Ki = and Ui = m g h At the end we have Kf = 21 m vf2 and Uf = Thus, according to Eq 6.51, we get: Ei = Ef ⇒ i.e., vf = Ki + Ui = Kf + Uf 2gh = ⇒ + m g h = 21 m vf2 + 2(9.8 m/s2 )(3 m) = 7.67 m/s (b) In the presence of a non-conservative frictional force, i.e Wnc = 0, then mechanical energy is not conserved and we can use Eq 6.56 to find the work done by friction on the boy as follows: Wnc ≡ Wf = Ef − Ei = (Kf + Uf ) − (Ki + Ui ) = ( 21 m vf2 + 0) − (0 + m g h) = 21 (30 kg)(5 m/s)2 − (30 kg)(9.8 m/s2 )(3 m) = −507 J Note that Wf is negative, since the work done by friction is negative i h fk N mg f Fig 6.16 Example 6.9 A block of mass m = kg is placed on a rough horizontal surface against a compressed spring with a spring constant of kH = 2,000 N/m The spring is compressed a distance of x = 20 cm, see Fig 6.17 The block is released, and then it moves to the right until it stops completely after rising onto a rough track of height h = 0.5 m Find the work done by friction 162 Work, Energy, and Power f Ei i =0 Ef Equilibrium pos ition =0 h x x=0 Fig 6.17 → Solution: As in Example 6.8, the normal force N does no work on the block since it is always perpendicular to each displacement element on the horizontal and curved parts of the track The only force that has a change in potential energy is the force of gravity Initially, we have Ki = and only an elastic potential energy 21 k x , i.e Ui = 21 k x At the end we have Kf = and only a gravitational energy m g h, i.e Uf = m g h In the presence of a non-conservative frictional force, Wnc = 0, the mechanical energy is not conserved We then use Eq 6.56 to find the work done by friction as follows: Wnc = Ef − Ei = (Kf + Uf ) − (Ki + Ui ) = (0 + m g h + 0) − (0 + + 21 kH x ) = m g h − 21 kH x = (2 kg)(9.8 m/s2 )(0.5 m) − 21 (2,000 N/m)(0.2 m)2 = −30.2 J Wnc is negative since the work done by friction is always negative 6.7 Conservation of Energy → When a block slides across a rough floor through a displacement d , the frictional → force f k (which is an external force) does work on the block, and we found from Example 6.7 that Wnc = Wf = −fk d This allows us to write the dissipated mechanical energy given by Eq 6.56 as follows: E= K+ U = −fk d (6.57) 6.7 Conservation of Energy 163 In fact, this dissipated energy is transferred as thermal energy to the block and the floor So, the energy of the block, which is considered to be our system, is not conserved When we expand our system to include both the block and the floor, the frictional force is no longer an external force, and the energy transfer will be within the system So, again we have an isolated system within which energy is conserved To find this conservation principle, we look at the decrease E in Eq 6.57 as the total amount of energy transferred as thermal energy to the block and floor If Eint represents the change in the thermal energy (which is an internal energy) of the system consisting of the block and floor, then we get: Eint = − E E+ which gives: Eint = K+ U+ (6.58) Eint = (6.59) This means that, although the mechanical energy of the block is not conserved, the sum of the mechanical energy of the block and the thermal energy of the block and floor is conserved This sum is called the total energy Etot of the block-floor system This conservation principle is called the law of conservation of energy and written as: Etot = K+ U+ conservation of energy Eint = for an isolated system (6.60) This law of conservation is not derived, but instead based on countless experiments done by scientists and engineers If the system is not isolated and applied external forces transfer energy to or from the system, then the work done on the system by external forces will be: W = Etot = K+ U+ Eint (non isolated system) (6.61) For example, in Fig 6.18, if we consider the rope to be external to the system, then the frictional force exerted by the rope on the metal rings of the system does an amount of work W on the system, transferring energy from the system to thermal energy in the rope while the values of K, U, and Eint change 164 Work, Energy, and Power Fig 6.18 A firewoman wrapping a rope around metal rings so that the rope rubs against the rings while she is descending from a helicopter Doing so, she will transfer energy from the gravitational potential energy of a system consisting of her, her gear, and the Earth to thermal energy gained by the rope and the rings While descending slowly, this allows most of the transferred energy to go to the rope and the rings rather than to her kinetic energy Example 6.10 A steel ball of mass m = g is projected vertically downward from a height h = 14.8 m with an initial speed v◦ = 10 m/s, see part a of Fig 6.19 The ball penetrates itself in sand to a depth d = 20 cm, see part c of the figure Neglect air resistance and take g to be 10 m/s2 (a) What is the change in the mechanical energy of the ball? (b) What is the change in the internal energy of the ball-Earthsand system? (c) What is the magnitude of the average force exerted by the sand on the ball in part b of the figure? Solution: (a) Let us take the reference point y = to be at the point where the ball stops completely, as shown in part c of the figure Therefore, at the stopping depth d, the kinetic energy and the potential energy are zero Thus: E = Ef − Ei = K+ U = (Kf − Ki ) + (Uf − Ui ) = (0 − 21 mv◦2 ) + (0 − mg[h + d]) = − 21 mv◦2 − mg(h + d) 6.7 Conservation of Energy 165 Inserting the given data into the final expression, we find: E = − 21 (5 × 10−3 kg)(10 m/s)2 − (5 × 10−3 kg)(10 m/s2 )(14.8 m + 20 × 10−2 m) Spring gun = −0.25 − 0.75 = −1 J y ti , Ei t,E o h t f ,Ef _ F Sand d x (a) (b) (c) Fig 6.19 (b) This system is isolated, and we can apply Eq 6.59 as follows: E+ Eint = or Eint = − E = −(−1 J) = J That is to say, as the ball moves through the sand, the sand exerts an upward force on the ball and thus dissipates all the mechanical energy of the ball, transforming it to thermal energy of the sand and ball (c) When the ball reaches the surface of the sand, its mechanical energy will be the same as the initial mechanical energy Ei , since air resistance is neglected