Work, Energy, and Power - HAPTEH Objectives By the end of...

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Unformatted text preview: HAPTEH Objectives By the end of this chapter you should be able to: - state the definition of work done by a force, W = F5 cos 9, appreciate the significance of the angle appearing in the formula and understand that this formula can only be used when the force is constant; - understand that the work done by a varying force is given by the area under the graph of force versus displacement; l . . - . n 2 - state the deflnltlons of kinetic energy, E k = gnu/2 (also Ek = 5?), i i i i i i i i l i a 3 3 i i 3 gravitational potential energy, E p = mgh, and elastic potential energy, a e — lkxz i e — 2 ' l - appreciate that gravitational potential energy can be calculated by measuring heights from an arbitrary level; - understand that, when frictional forces are absent, the total energy 3 E : Ek + Ep + Ee = firm/2 + mgh + %kx2 is conserved; i - use the work-kinetic energy relation that states that the work done by 3 the net force is the change in kinetic energy; i - understand that. in the presence of external forces, the work done is the 3 change in the mechanical energy, W = A5; 3 - state the definition of power, P = %, and its very useful form in i mechanics, P 2 FV; % - calculate the efficiency of simple machines; i - understand that in all collisions momentum is always conserved, but that kinetic energy is only conserved in elastic collisions. 100 Core - Mechanics Figure 7.1 A force moving its point of application performs work. Figure 71 _ ,_ _ ‘ I 7 angle between the force and any small -‘ ~1“as”sew-.ee'eesewwfgsmsweetwwesg ' _ . ' displacement of the mass as It revolves around the ' ‘ l? f circle is 90° and since cos 90° 2 0, the work ' I done is zero. Q3 A force of constant magnitude 25 N acts on a body that moves along a curved path. The direction of the force is along the velocity vector of the body, i.e. it is tangential to the path. Find the work after the mass moves a distance of 50 m along the curved path. oommmgrwms. .. m, w- Example question Q1 . A mass is being pulled along a level road by a Answer rope attached to it in such a way that the rope The diagram on the left of Figure 7.3 ShOWS the makes an angle of 400° with the horizontal. The path of the body. The diagram on the right is an force in the rope .is 20.0 N. What is the work enlargement of part of the path. We see that the " done by this force in moving the mass a distance work done when the body travels a small distance of 8.00 m along the level road? AS is Answer FAs cos 0° 2 FAs Applying the definition of work done, we have Breaking up the entire path into small bits in this W: F5 COS 6 way, and adding the work done along each bit, we find that the total work done Is W :Fs, where F : 20.0 N, 57: 8.00 m and 6 = 40". Thus where S is the distance travelled along the W: 20 X 8 X COS 40° CUrVEd path' 2 123 J If the force is not constant or the motion does F not take place in a straight line, or both, we for must be careful. First consider the case of a force of constant magnitude when the motion Figure 7_3_ is not along a straight line. If the force is not constant in magnitude, we Example quesnons must be supplied with the graph that shows the Q2 _ variation of the magnitude of the force with Flnd the work done by the tense“ m a strmsr as a distance travelled. Then we have the following mass attached to the end of the string performs important result; circular motion (see Figure 7.2). . \§,=¢yygx-§§=W-wfit=-§>W~WWW_WWVWWW_WW . . '1' .. _~ . ,- e w» y ewetfi - , ~ ‘ ‘ 3 rates Answer gm” 3% ‘ n} ._ 3_._ z . wwfiia “9&1 ' . , eeewemewe J3“. . This Is a case where the force, although constant 15$. fiWflfi-‘aN-fiw‘w in magnitude, changes in direction. However, the assessme- 2.7 Work, energy and power 101 _—_ We can apply this result to the case of the tension in the spring. Since T : kx, where k is the spring constant and X the extension (or compression) of the string, the graph of force versus position is as shown in Figure 7.4. extension Figure 7.4 The area under a force—distance graph gives the total work performed by the non— constant force. To find the work done in extending the spring from its natural length (X = 0) to extension X, we need to calculate the area of the triangle whose base is x and height is T = RX . Thus area = %kx xx _1 2 _ka The work to extend a Spring from its natural length by an amount X is thus W = %kx2 It follows that the work done when extending a spring from an extension X1 to an extension X2 (so x2 > 'x1) is W: iHXi-Xi) Work done by gravity We will now concentrate on the work done by a very special force, namely the weight of a mass. Remember that weight is mass times acceleration due to gravity and is directed vertically down. Thus, if a mass is displaced horizontally, the work done by mg is zero, since in this case the angle is 90°: W = mgd cos 900 = 0. (See Figure 7.5.) displacement Figure 7.5 The force of gravity is normal to this horizontal displacement so no work is being done. " ' Note that we are not implying that it is the Weight that is forcing the mass to move along the table; We are calculating the work done by a particular force (the weight) if the mass (somehow) moves in a particular way. If the body falls a vertical distance h, then the work done by W is +mgh. The force of gravity is parallel to the displacement, as in Figure 7.6a. If the mass is thrown vertically upwards to a height h from the launch point, then the work done by W is —mgh since now the angle between direction of force (vertically down) and displacement (vertically up) is 180°. The force of gravity is parallel to the displacement but opposite in direction, as in Figure 7.6b. displacement (a) (b) Figure 7.6 The force of gravity is parallel to the displacement in (a) and anti-parallel (i.e. parallel but opposite in direction) in (b). 102 Core — Mechanics —————_u____ Consider now the case where a mass moves along some arbitrary path, as shown in Figure 7.7. start Figure 7.7 The work done by gravity is independent _of the path followed. The path consists of horizontal and vertical segments. We now ask about the work done by the weight of the mass. The work done by my will be equal to the sum of the work done along each horizontal and vertical step. But my does no work along the horizontal steps since the angle between the force and the displacement in that case will be 900 and cos 90° 2 0. We are thus left with the vertical steps only. The work done along each step will be :lzmg Ah, where Ali is the step height. The plus sign is used when we go down a step and the minus sign when we go up a step. (In Figure 7.7 the mass will be forced to go up twice and down eight times.) Thus, what counts is the net number of steps going down (six in our figure). But, this adds up to the vertical distance separating the initial and final position. Hence, the work done by mg is mgh. If the start and finish positions are joined by an arbitrary smooth curve rather than a ‘staircase’, the result is still the same. This is because we can always approximate a smooth curve by a series of horizontal and vertical steps; the quality of the approximation depends on how small we take the steps to be. This means that: shawlzéffswfisgemwgwfsww «WWW _ mm. ,w t ‘ 3% 3% "é as as as as {6 Work done in holding something still If you try to hold up a heavy object, such as a chair, you will soon get tired. However, the force with which you are holding the chair does zero work since there is no displacement. This is somewhat unexpected. We normally associate getting tired with doing work. Indeed, the forces inside the muscles of the arm and hand holding the chair do work. This is because the muscles stretch and compress and that requires work, just as stretching and compressing a spring does. As we just saw, the weight mg of a mass m a height h from the ground will perform work mgh if this mass moves from its position down to the ground. The ability to do this work is there because the mass just happens to be at a height h from the ground. The ability to do work is called energy. When the force in question is the weight (which depends on gravity), we call this energy gravitational potential energy: Ep =mgh Any mass has gravitational potential energy by virtue of its position. But what determines h? Obviously, we have to choose a reference level from which we will measure heights. But we can choose any level we like. A mass m z 2 kg sitting on a table 1 In from the floor will have EFl = 2 x 10 x l = 20 J if the reference level is the floor, but will have E p = 0 if the reference level is the table. If the reference level is chosen to be the ceiling, 2 m above the table, then Ep = “2 x 10 x 2 = —40 J. (See Figure 7.8.) 2m Figure 7.8 The mass has different potential energies depending on the reference level chosen. 2.7 Work, energy and power 103 m———— 80, the same mass will have different gravitational potential energy depending on What reference level we choose. This might seem to make E p a useless quantity. But if you are patient, you will see that this is not the case. Potential energy can be understood in the following way. Consider a mass resting on a horizontal floor. If an external force equal to mg is applied to the mass vertically up and the mass moves without acceleration to a position it metres higher than the floor, the work done by the external force is mgh. What has become of this work? This work has gone into gravitational potential energy of the mass. This energy is stored as potential energy in the new position of the mass. Similarly, if a spring is initially unstretched and an external force stretches it by an amount x, then the work done by this external force is gkxz. This work is stored as elastic potential energy in the (now stretched) spring. mew “3e easements r waeW-nezfifie. ,saisuwneawme ew'aknwnaéaewaesaisv - tenppi new earnestnesaaniatwam i ‘ , firearms-teammate fishermen: m: - ass. , faststeataamai __ eaiggégggfigameweeefiefizneg the ‘ - $§$¥§£§§§m nenagpetseansaaaagana weeyemewsswew $549 {ewe awrmwwwmsw warms» sake-v35 . .. .. - - fie! wfiflfim we Ihifi=fifiW$i§Wfifi$mfifi Mags; :eflmwfi. ’W'.‘ &=%%&W§>Wéfivmwm five Wsfifimemwmaw ewwemm mmmgeawwfim «m.w.mw a. Example question Q4 A mass of 10 kg rests on top of a vertical spring whose base is attached to the floor. The spring compresses by 5 cm. What is the spring constant of the spring? How much energy is stored in the spring? Answer The mass is at equilibrium so mgzkx =>k=fl x _E _0.05 = 2000 N m’1 The stored energy is l i kx2 E32 1 E x 2000 (0.05)2 2.5] The work—kinetic energy relation What effect does the work done have on a body? When a body of mass m is acted upon by a net force F, then this body experiences an acceleration a = f; in the direction of F. Suppose that this body had speed V0 when the force was first applied to it and that the speed after moving a distance x (in the direction of the net force) becomes Vf, as shown in Figure 7.9. # V0 F Vf ‘ Figure 7.9 A force accelerates a mass, increasing its kinetic energy. We know from kinematics that V3: V5 + 20x so replacing the acceleration by F/m we find F 2 2 V = V + 2— X r 0 m => Fx 2 Emvf — Emvo But Fx is the work done on the mass. This work equals the change in the quantity E k = gmvz, a quantity called the kinetic energy of the mass. We thus see that the net work done on a mass results in a change of the kinetic energy of the object. This is a very useful result with applications in many areas of physics. i wwwww—Wwwwwwwagww .tfifiififi“$§i§§§%§$ 104 Core - Mechanics Example questions A mass of 5.00 kg moving with an initial velocity of 12.0 m s’1 is brought to rest by a horizontal force over a distance of 12.0 m. What is the force? Answer The change in the kinetic energy of the mass is (final minus initial) 1 1 0— —mv2 = —— x 5.00 x144 2 2 =~36OJ The work done by the force f is — f5 = —1 2 1‘ Hence —12f=—360 :f:30.0N (There is a minus sign in the work done by f because the force is acting in a direction opposite to the motion and cos 180° = —1.} Q6 An electron is acted upon by an electric force that accelerates it from rest to a kinetic energy of 5.0 x “IO”19 J. If this is clone over a distance of 3.0 cm, find the electric force. Answer The work done by the electric force is the change in kinetic energy (the electric force is the only force acting and so it is the net force) and equals the product of force times distance. Hence F x 0.03 25.0 x 10—‘9 _ => F=i.7x10’17N Q7 A mass m hangs from two strings attached to the ceiling such that they make the same angle with the vertical (as shown in Figure 7.10). The strings are shortened very slowly so that the mass is raised a distance h above its original position. What is the work done by the tension in each string as the mass is so raised? Figure 7.10. Answer The net work done is zero since the net force on the mass is zero. The work done by gravity is 7mg!) and thus the work done by the two equal tension forces is +mgh. The work done by each is thus mgh/Z. Q8 A mass m hangs vertically at the end of a string of length L. A force F is applied to the mass horizontally so that it slowly moves to a position a distance h higher, as shown in Figure 7.11. What is the work done by the force F? (Note: F is not constant.) I Figure 7.11. Answer The answer is obtained at once by noting that since the mass is in equilibrium at all times the net force is zero and hence the work done by the net force is zero. But T does zero work since it is always normal to the direction in which the mass moves (along the arc of a circle). The weight does work imgh and thus the work done by F must be +mgh. Conservation of energy We have already seen that, when a net force F performs work W on a body, then the kinetic energy of the body changes by W (here the 2.7 Work. energy and power 105 ___—_——____.—.——_-— subscripts ‘i’ and ‘f’ stand for ‘initial’ and ‘final') W = AEk = §mvf2 — émviz Consider now the case where the only force that does work on a body is gravity. This corresponds to motions in which the body is either in free fall (gravity is the only force acting) or the body is sliding on a frictionless surface (we now have the normal reaction force acting here as well as gravity but this force does zero work since it is normal to the direction of motion). Suppose that the vertical height of the body when the velocity is V5 is H, and the velocity becomes v; at a height of h from the reference level (see Figure 7.12). Figure 7.12 The total energy at the top and bottom (and at any point in between) of the incline is the same. The work done by gravity is simply mg(H — h) and so man—h) = imvr — imvr kinetic only Figure 7.13. l This can be rearranged as 2 mgh + %mva = mgH + %mvi which shows that in the motion of this body the sum of the kinetic and potential energies of the mass stays the same. Calling the quantity nigh + %mv2 the total energy of the mass, then the result we derived states that the total energy E of the mass stays the same at all times. i.e. the total energy is conserved, As the mass comes down the plane, its potential energy decreases but its kinetic energy increases in such a way that the sum stays the same. We have proven this result in the case in which gravity is the only force doing work in our problem. Consider now the following example questions. Example questions Q9 Find the speed of the mass at the end of a pendulum of length 1.00 m that starts from rest at an angle of 10° with the vertical. Answer Let us take'as the reference level the lowest point of the pendulum (Figure 7.13). Then the total energy at that point is just kinetic, Ek = 1imvz, where v is the unknown speed. At the initial point the total energy is just potentiai, E9 = mgh, where 106 Core — Mechanics W h is the vertical difference in height between the two positions, that is h: 1.00 — 1.00cos10" =0.015m [see Figure 7.13) Thus 1 2 _ imv _ mgh v= 2gb = 0.55 ms" Note how the mass has dropped out of the problem. (At positions other than the two shown, the mass has both kinetic and potential energy.) Q10 A mass rolls up an incline of angle 28° with an initial speed of 3.0 m 571. How far up the incline will the mass get? Answer Let the furthest the mass will get be a height h from the floor, as shown in Figure 7.14. At this point the kinetic energy must be zero since otherwise the mass would have climbed higher. Then the total energy at this point is just E = mgh. The total energy at the initial position is E = 1imv2 and so h = V_2 23 giving h : 0.45 m. The distance moved along the plane is thus potential kinetic only - only Figure 7.14. Q11 What must the minimum speed of the mass in Figure 7.15 be at the initial point such that the mass makes it over the barrier of height h? Figure 7.15. Answer To make it over the barrier the mass must be able to reach the highest point. Any speed it has there will then make it roll over. Thus, at the very least, we must be able to get the ball to the highest point with zero speed. Then the total energy would be just E : mgh and the total energy at the starting position is E = §mv2. Thus, the speed must be bigger than v = Note that if the initial speed u of the mass is larger than v 2 Jig—H, then when the mass makes it to the other side of the barrier its speed will be the same as the starting speed u. Q12 - . A mass rolls off a 1.0 m high table with a speed of 4.0 m 5*, as shown in Figure 7.16. With what speed does it strike the floor? Figure 7.16. Answer The total energy of the mass is conserved. As it leaves the table it has total energy given by E = 1Emv2 + mgh and as it lands the total energy is E = gmu2 (u is the speed we are looking for]. Equating the two energies gives 1Emu2 = 32—mv2+ mgh =>u2= v2+2gh =l6+20=36 =>u = 6.0 m s—1 l l I i i Q13 A ball is thrown vertically upward with a speed of 4.0 m 571 from a height of 1.0 m from the floor, as shown in Figure 7.17. With what speed does the ball strike the floor? i 4.0 m s—1 1.0m Figure 7.17. Answer Working in precisely the same way as in the previous example we find —mu2 = %mv2 + mgh i :U22V2-i-2gh =l6+20 =36 =>u = 6.0 ms"1 (The answer is the same as that for example question 12 — why?) If, in addition to the weight, there are spring tension forces acting in our system, then the previous discussion generalizes to again lead to E1=Ef where now the total mechanical energy includes elastic potential energy as well, that is E = %mV2 +mgh + got2 Example question A body of mass 0.40 kg is held next to a compressed spring as shown in Figure 7.18. The spring constant is k 2 250 N m‘1 and the compression of the spring is 12 cm. The mass is then released. Find the speed of the body when it is at a height of 20 cm from the horizontal. 2.7 Work, energy and power 107 l l 20 cm spring Figure 7.18. Answer Initially the total mechanical energy of the system is only the elastic potential energy of the spring 1 2 mechanical energy consists of kinetic and gravitational energies only. The spring has decompressed and so has no elastic potential energy. Then E = limvz + mgh = grow + (0.4)(10)(0.20) = 0.2 v2 + 0.8 Thus 0.2v2+0.8= 1.8 =>0.2v2 z 1.0 =>V2z5 =>v= 2.2ms‘1 Frictional forces In the presence of friction and other resistance forces, the mechanical energy of a system (i.e. the sum of kinetic, gravitational potential and elastic Potential energies} will not be conserved. These forces will, in general, decrease the total mechanical energy of the system. Similarly, external forces. such as forces due to engines, may increase the mechanical energy of a system. In these cases we may write W=AE l l —1§>< 250 x 0.122 = 1.8] At a height of 20 cm from the floor, the total I l 108 Core - Mechanics —————————_—_—_—_ where W stands for the total work done by the external forces and AE is the change in the mechanical energy of the system. By external forces we mean forces other than weight and spring tension forces. This equation is easily understood in the following way. If there are no external forces, then W = 0, AE = 0 and the total mechanical energy stays the same: it is conserved. If, on the other hand, external forces do act on the system, then the work they do goes into changing the mechanical energy. If the work done is negative (resistance forces), the mechanical energy decreases. If the work done is Positive (pulling forces), the mechanical energy increases. Example question A body of mass 2.0 kg (initially at rest) slides down a curved path of total length 16 m as 5h0wn in Figure 7.19. When it reaches the bottom, its speed is measured and found to equal 6.0 m s“1 . Show that there is a force resisting the motion. Assuming the force to have constant magnitude, determine what that magnitude is. Figure 7.19. Answer Without any resistance forces, the speed at the bottom is expected to be v=‘/2gh =v2 x '10 x 5.0 :10ms‘1 The measured speed is less than this and so there is a resistance force. The total mechanical energy at the top is Etop : mgh = 2.0 x10 x 5.0 = 100] At the bottom it is 1 2 IEbottorn = imv 2 % X 2.0 x (6.0)2 :36] The total energy decreased by 64 J — this must be the work done by the resistance force. Thus f5=64J 64 =>f=— 16 =4.0N We have seen that in the presence of external forces the total mechanical energy of a system is not conserved. The change in the total energy is the work done by the external forces. Another way of looking at this is to say the change in the mechanical energy has gone into other forms of energy not included in the mechanical energy, such as thermal energy (‘heat’) and sound. In this way total energy (which now includes the other forms as well as the mechanical energy) is conserved. This is the general form of the law of conservation of energy, one of the most important principles of physics. waste -. - isms fies as: Power When a machine performs work, it is important to know not only how much work is being done but also how much work is performed within a given time interval. A cyclist performs a lot of work in a lifetime of cycling, but the same work is performed by a powerful car engine in a much shorter time. Power is the rate at which work is being performed. 2.7 WorkI energy and power 109 Another common unit for power when it comes to machines and car engines is the horsepower, hp, a non-SI unit that equals 746 W Consider a constant force F , which acts on a body of mass m. The force does an amount of work F Ax in moving the body a small distance Ax along its direction. If this work is performed in time At, then AW P =—— At Ax At EFV where V is the instantaneous speed of the mass. This is the power produced in making the body move at speed V. As the speed increases, the power necessary increases as well. Consider an aer0plane moving at constant speed on a straight- line path. If the power produced by its engines is P, and the force pushing it forward is F, then P, F and V are related by the equation above. But since the plane moves with no acceleration, the total force of air resistance must equal F. Hence the force of air resistance can be found simply from the power of the plane’s engines and the constant speed with which it coasts. Example questions Q16 .. What is the minimum power required to lift a mass of 50.0 kg up a vertical distance of 12 m in 5.0 5? Answer The work performed to lift the mass is mgh:50.0 x10 x12 26.0x103l The power is thus 6.0x103 —=1200w This is only the minimum power required. in practice, the mass has to be accelerated from rest, which will require additional work and hence more power. Q17 A helicopter rotor whose length is R pushes air downwards with a speed v. Assuming that the density of air is constant and equals ,0 and the mass of the helicopter is M, find v. You may assume that the rotor forces the air in a circle of radius R (spanned by the rotor) to move with the downward speed v. Hence find the power developed by the engine. How does this power depend on the linear size of the helicopter? Answer The momentum of the air under the rotor is mv, where m is the mass of air in a circle of radius R. In time At the mass is enclosed in a cylinder of radius R and height vAt. Thus, the momentum of this mass is anZ vat and its rate of change is an2 v2. This is the force on the helicopter upwards, which must equal the helicopter’s weight of Mg. Thus Mg 2 prrR2 v2 Mg 3} = v pn'l-Z’2 The power required from the helicopter engine is thus PFv Ms pr: R2 Mg To find the dependence on a typical linear size L of the helicopter, note that the weight depends on L as L3 and so L3 P or twp or L"2 This implies that if the length of a helicopter is 16 times that of a model helicopter, its required 1 10 Core - Mechanics h—_—__—— power will be 167/2 % 16000 times larger than that for the model. Efficiency Suppose that a mass is being pulled up along a rough inclined plane with constant Speed. Let the mass be 15 kg and the angle of the incline 45°. The constant frictional force opposing the motion is 42 N. Figure 7.20. The forces on the mass are shown in Figure 7.20 and we know that R :mgcose = 106.1 N F =mgsin6 + f = 106.1 +42 3 148.1 N @150N since the mass has no acceleration. Let the force raise the mass a distance of 20 In along the plane. The work done by the force F is W = 148.1 x 20 = 2960 J a 3.0 x103J The force effectively raised the 15 kg a vertical height of 14.1 In and so increased the potential energy of the mass by mgh = 2121 J. The efficiency with which the force raised the mass is thus useful work ll 2 actual work w' 2121 T m = 0.72 Example question A 0.50 kg battery-Operated toy train moves with constant velocity 0.30 m 5'1 along a level track. The power of the motor in the train is 2.0 W and the total force opposing the motion of the train is 5.0 N. (a) What is the efficiency of the train’s motor? (b) Assuming the efficiency and the opposing force stay the same, calculate the speed of the train as it climbs an incline of 10.00 to the horizontal. Answer (a) The power delivered by the motor is 2.0 W. Since the speed is constant, the force developed by the motor is also 5.0 N. The power used in moving the train is Fv = 5.0 x 0.30 = 1.5 W. Hence the efficiency is 1.5 W Zr 2.0 w = 0.75 7? (b) The net force pushing the train up the incline is F = mgsinQ +5.0 =0.50 x 10 x sin10°+5.0 : 5.89 N as 5.9 N Thus 5.89 x v n 2 2.0 w = 0.75 2.0 x 0.75 => v: 5.89 = 0.26 ms" Kinetic energy and momentum We have seen in Chapter 2.6 on momentum that, in a collision or explosion where no external forces are present, the total momentum of the system is conserved. You can easily convince yourself that in the three collisions illustrated in Figure 7.21 momentum is conserved. The incoming body has mass 8.0 kg and the other a mass of 12 kg. 2.7 Work, energy and power 1 l 1 l _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ “ ' ~ _ _ __ i _ _ ‘ T ‘ - _ _ _ T _ _ _ _ _ _ _ _ _ _ _" l T m u _ _ _ _ _ _ _ n u _ " M _ _ _ _ _ __ r 10 ms” 10 m s-1 before after Figure 7.2.1 Momentum is conserved in these three collisions. In the first collision the bodies have stuck together and move as one. In the second the incoming body has slowed down as a result of the collision and the heavy body moves faster. In the third the incoming .body has bounced back. Let us examine these collisions from the point of View of energy. In all cases the total kinetic energy before the collision is Ek=%x8xl02=400J The total kinetic energy after the collision in each case is: casel Ek:l§x20_x42=l6OJ case2 Ek=%x8x12+%x12x62=220l case3 Ek=%x8x22+%x12'x82:400l We thus observe that whereas momentum is conserved in all cases, kinetic energy is not. When kinetic energy is conserved (case 3), the collision is said to be elastic. When it is not (cases 1 and 2), the collision is inelastic. In an inelastic collision, kinetic energy is lost. When the bodies stick together after a collision (case 1), the collision is said to be totally inelastic and in this case the maximum possible kinetic energy is lost. The lost kinetic energy gets transformed into other forms of energy, such as thermal energy, deformation energy (if the bodies are permanently deformed as a result of the collision) and sound energy. Example questions Qw A moving body of mass m collides with a stationary body of double the mass and sticks to it. What fraction of the original kinetic energy is lost? ' Answer The original kinetic energy is 1Emvz, where v is the speed of the incoming mass. After the collision the two bodies move as one with speed u that can be found from momentum conservation: mv = (m + 2m)u am e 3 T 3 The total kinetic energy after the collision is therefore (gr—"e and so the lost kinetic energy is 2 The fraction of the original energy that is lost is thus mv2/3 mv2/2 _ 2 3 Q20 A body at rest of mass M explodes into two pieces of masses M/4 and 3M/4. Calculate the ratio of the kinetic energies of the two fragments. l l 2 Core — Mechanics Answer Here it pays to derive a very useful expression for kinetic energy in terms of momentum. Since mv2 Ek= —_ 2 it follows that mv m Ek=¥xfi m 2m = % The total momentum before the explosion is zero, so it is zero after as well. Thus, the two fragments must have equal and opposite momenta. Hence Eight _ lot/(might) Eheavy Mheavy) ll The problem of least time In Figure 7.22 a number of paths join the starting position A to the final position B. B Figure 7.22. A mass m at A will start with a tiny speed and move down to B. As We saw, the speed at B will be the same no matter what path the mass follows. The Speed will equal J29? in all cases. This does not mean, however, that the time taken is the same for all paths. Finding the path joining A to B such that a mass takes the least ——-—_—_—_——_~__ time is a famous problem in physics and requires the development of a branch of calculus called the calculus of variations. It is called the brachistochrone (least time) problem. The answer is that the curve joining A to B must be a cycloid. This is the curve traced out by a point on the rim of a wheel as the wheel rolls. In Figure 7.22, curve IV resembles a cycloid most. This problem was posed to both Newton and Leibniz (the inventors of calculus) by the Swiss mathematician Bernoulli. When Bernoulli saw the solutions given by the two men, he is supposed to have said of Newton ‘one can always tell a lion by its claws’. 1 A horizontal force of 24 N pulls a body a distance of 5.0 m along its direction. Calculate the work done by the force. 2 A block slides along a rough table and is brought to rest after travelling a distance of 2.4 m. The frictional force is assumed constant at 3.2 N. Calculate the work done by the frictional force. 3 A block is pulled as shown in Figure 7.23 by a force making an angle of 20° to the horizontal. Find the work done by the pulling force when its point of application has moved 15 m. Figure 7.23 For question 3. 4 A block of mass 4.0 kg is pushed to the right by a force F = 20.0 N. A frictional force of 14.0 N is acting on the block while it is moved a distance of 12.0 m along a horizontal floor. The forces acting on the mass are shown in Figure 7.24. (a) Calculate the work done by each of the four forces acting on the mass. (in) Hence find the net work done. -(c] By how much does the kinetic energy of the mass change? 2.7 Work, energy and power 1 1 3 Figure 7.24 For question 4. 5 A weight lifter slowly lifts a 100 kg mass from the floor up a vertical distance of 1.90 m and then slowly lets it down to the floor again. (a) Find the work done by the weight of the mass on the way up. (b) Find the work done-by the force exerted by the weight lifter when lifting the weight up. (c) What is the total work done by the weight on the way up and the way down? 6 A block of mass 2.0 kg and an initial speed of 5.4 rn 5"1 slides on a rough horizontal surface and is eventually brought to rest after travelling a distance of 4.0 m. Calculate the frictional force between the block and the surface. 1is 7 A spring of spring constant k = 200 N m— slowly extended from an extension of 3.0 cm to an extension of 5.0 cm. How much work is done by the extending force? 8 A spring of spring constant k: 150 N m“ is compressed by 4.0 cm. The spring is horizontal and a mass of 1.0 kg is held to the right end of the spring. If the mass is released, with what speed wiil it move away? ' 9 Look at Figure 7.25. (a) What is the minimum speed vthe mass must have in order to make it to position B? What speed will the mass have at B? (b) If v: 12.0 m 5", what will the speed be at A and B? Figure 7.25 For question 9. 10 A mass is released from rest from the position shown in Figure 7.26. What will its speed be as it goes past positions A and B? Figure 7.26 For question 10. 11 The speed of the 8.0 kg mass in position A in Figure 7.27 is 6.0 m 5". By the time it gets to B the speed is measured to be 12.0 m 571. Figure 7.27 For question 11. What is the frictional force opposing the motion? (The frictional force is acting along the plane.) of spring constant 12.0 N m“ decompreSSes. The amount of compression is 10.0 cm (see Figure 7.28). With what speed does the ball exit the gun, assuming that there is no friction between the ball and the gun? if, instead, there is a frictional force of 0.05 N opposing the motion of the ball, what will the exit speed be in this case? 4—-—~—-—> 10 cm Figure 7.28 For question 12. 13 A variable force F acts on a body of mass m = 2.0 kg initialiy at rest, moving it along a straight horizontal surface. For the first 2.0 m the force is constant at 4.0 N. in the next 2.0 m it is constant at 8.0 N. in the next 2.0 m it drops from 8.0 N to 2.0 N uniformly. It then 12 A toy gun shoots a 20.0 g ball when a spring I l l increases uniformly from 2.0 N to 6.0 N in the 1 14 Core — Mechanics “——-—-——————_—.— 14 15 16 17 13 19 next 2.0 m. It then remains constant at 6.0 N for the next 4.0 m. (a) Draw a graph of the force versus distance. (b) Find the work done by this force. (C) What is the final speed of the mass? A body of mass 12.0 kg is dropped vertically from rest from a height of 80.0 m. Ignoring any resistance forces during the motion of this body, draw graphs to represent the variation with distance fallen of (a) the potential energy; (b) the kinetic energy. For the same motion draw graphs to represent the variation with time of (C} the potential energy; (d) the kinetic energy. (e) Describe qualitatively the effect of a constant resistance force on each of the four graphs you drew. A 25.0 kg block is very slowly raised up a 7 vertical distance of 10.0 rn by a rope attached I to an electric motor in a time of 8.2 5. What is the power developed in the motor? The engine of a car is developing a power of 90.0 kW when it is moving on a horizontal road at a constant speed of 100.0 km th. What is the total horizontal force opposing the motion of the car? The motor of an elevator develops power at a rate of 2500 W. (a) At what speed can a 1200 kg ioad be raised? (b) In practice it is found that the load is lifted more slowly than indicated by your answer to (a). Suggest reasons why this is so. A load of 50.0 kg is raised a vertical distance of 15 m in 125 s by a motor. (a) What is the power necessary for this? (b) The power supplied by the motor is in fact 80 W. Calculate the efficiency of the motor. (c) If the same motor is now used to raise a load of 100.0 kg and the efficiency remains the same, how long would that take? 7 For cars having the same shape but different size engines it is true that the power developed by the car’s engine is 20 21 22 23 proportional to the third power of the car’s maximum speed. What does this imply about the speed dependence of the wind resistance force? The top speed of a car whose engine is delivering 250 kW of power is 240 km W. Calculate the value of the resistance force on the car when it is travelling at its top speed on a level road. Describe the energy transformations taking place when a body of mass 5.0 kg: - (a) falls from a height of 50 m without air resistance; (b) falls from a height of 50 m with constant speed; (c) is being pushed up an incline of 30° to the horizontal with constant speed. An elevator starts on the ground floor and stops on the 10th floor of a high-rise building. The elevator picks up a constant speed by the time it reaches the 1st floor and decelerates to rest between the 9th and 10th floors. Describe the energy transformations taking place between the 1st and 9th floors. A car of mass 1200 kg starts from rest, accelerates uniformly to a speed of 4.0 m s—1 in 2.0 s and continues moving at this constant speed in a horizontal straight line for an additional 10 s. The brakes are then applied and the car is brought to rest in 4.0 s. A constant resistance force of 500 N is acting on the car during its entire motion. (a) Calculate the force accelerating the car in the first 2.0 s of the motion. (bl Calculate the average power developed by the engine in the first 2.0 s of the motion. (c) Calcuiate the force pushing the car forward in the next 10 s. (d) Calcuiate the power developed by the engine in those 10 s. (e) Calculate the braking force in the last 4.0 s of the motion. (fl Describetheenergy,transformationsthat, have taken place in the 16 s of the motion of this car. 24 A mass of 6.0 kg moving at 4.0 m s—1 collides with a mass of 8.0 kg at rest on a frictionless surface and sticks to it. How much kinetic energy was lost in the collision? 25 Two masses of 2.0 kg and 4.0 kg are held in place, compressing a spring between them. When they are released, the 2.0 kg moves away with a speed of 3.0 m s’l. What was the energy stored in the spring? 26 A block of mass 0.400 kg is kept in place so it compresses a spring of spring constant 120 N m‘1 by 15 cm (see Figure 7.29). The block rests on a rough surface and the frictional force between the block and the surface when the block begins to slide is 1.2 N. Figure 7.29 For question 26. (a) What speed will the block have when the spring returns to its natural length? (b) What percentage is this of the speed the mass would have had in the absence of friction? 27 Two bodies are connected by a string that goes over a pulley, as shown in Figure 7.30. , The lighter body is resting on the floor and the other is being held in place a distance of 5.0 m from the floor. The heavier body is then released. Calculate the speeds of the two bodies as the heavy mass is about to hit the floor. Figure 7.30 For question 27. __._\____..._ 28 29 30 31 2.7 Work, energy and power 1 ‘l 5 _—_——.——__ A mass m of 4.0 kg slides down a frictionless incline of6 = 30° to the horizontal. (a) Plot a graph of the kinetic and potential energies of the mass as a function of time. (b) Plot a graph of the kinetic and potential energies of the mass as a function of distance travelled along the incline. (c) On each graph, plot the sum of the potential and kinetic energies. The mass starts from rest from a height of 20 m. Show that an alternative formula for kinetic energy is Ek = %, where p is the momentum of the mass m. This is very useful when dealing with collisions. A body of mass M, initially at rest, explodes and splits into two pieces of mass M/3 and 2M/3, respectively. Find the ratio of the kinetic energies of the two pieces. (Use the formula'from the previous problem.) A mass m is being pulled up an inclined-plane of angle 9 by a rope along the plane. (a) What is the tension in the rope if the mass moves up at constant speed v? (b) What is the work done by the tension when the mass moves up a distance of d m along the plane? (c) What is the work done by the weight of the mass? (d) What is the work done by the normal reaction force on the mass? (e) What is the net work done on the mass? 33‘ A car of mass 1200 kg is moving on a horizontal road with constant speed 30 m s". The engine is then turned off and the car will eventually stop under the action of an air 1 16 Core — Mechanics resistance force. Figure 7.31 shows the (a) the speed of the jumper when she has variation of the car’s speed with time after the fallen by 12 m; engine has been turned off. (b) the maximum speed attained by the jumper during her fall. (c) Explain why the maximum speed is reached after falling more than a distance of 12 m (the unstretched length of the rope). 36 A carriage of mass 800 kg moving at 5.0 m s‘1 collides with another carriage of mass 1200 kg that is initially at rest. Both carriages are equipped with buffers. The graph in Figure 7.32 shows the velocities of the two Figure 7.31 For question 33. (a) Calculate the average acceleration of the car in the first and second 10 s intervals. (b) Explain why it takes longer to reduce the WW. M... carriages before, during and after the collision. sagas». _ “wit? -_ gfig speed from 20.0 m s"3 to 10.0 m s‘1 at"??? W- sta- fifié W -- We: compared with from 30.0 m s’1 to 20.0 m 5—1. (C) The average speed in the first “IO 3 interval is 21.8 m s’1 and in the second it is 13.5 m 5". Use this information and your answer in (a) to deduce that the air resistance force is proportional to the square of the speed. (d) Calculate the distance travelled by the car Figure 7.32 For question 36. in the first and second 10 5 intervals. (e) Calculate the work done by the resistance force in the first and second 10 5 intervals. (3) Show that the COillSEOH has been ElaSth} (b) calculate the average force on each carriage during the collision; (c) calculate the impulse given to the heavy carriage. ~ If the buffers on the two carriages had been stiffer, the time of contact would have been Fess but the final velocities would be unchanged. How would your answers to (b) and (c) change? (e) Calculate the kinetic energy of the two carriages at the time during the collision when both have the same velocity and compare your answer with the final kinetic 35 For the bungee jumper of mass 60 kg in energy of the carriages. How do you question 34, calculate: account for the difference? Use the graph to: 34 A bungee jumper of mass 60 kg jumps from a bridge 24 m above the surface of the water. The rope is 12 m long and is assumed to obey Hooke’s law. (on (a) What should the spring constant of the rope be if the woman is to just reach the water? (b) The same rope is used by a man whose mass is more than 60 kg. Explain why the man will not stop before reaching the water. (Treat the jumper as a point and ignore any resistance to motion.) Vad—bf‘n C‘Qw 31w?flimstatitifiany;p rt] “” .W-MxmuzM-vM‘W‘W" A. .35. prfith‘fei-O‘Fi www'bnws“? ‘Nfimbofla w :5 no Vsflupaanm 0'10 »\ "ac 4 .5; x' )- wig mv'a" :L ‘H yeloeitlhe “My; V ' . - Q Hymaij 43";1w-W§t _4. www‘“ Hwy“; may? ' _.‘ ‘- '. 5‘9"" ’“prww F51: be“). a. Wmmwxwwu :géawfi’fjb », ‘ V‘ v.5? \4 «ak‘W-“A‘” . . "w"? urwémxxwx W v.94)”, tvvfifl e:2.333E9r;q11esti‘ofi 4w “aw-Tare: «my . Wm W e Wha-twq;n' fig?! . 4» __ A «w .afierécojhs 2m: 8%? I W 'I W ,4 mw mom Zffi’étgén. nitral rr 93w is“? "a % .E?fo»$.'%«?wfirkm«6 33's: 43 A spring of natural length 0.150 m and spring constant 4.00 N m‘1 is attached at point P, as shown in Figure 5.26. The other end of the spring is attached to a ring that goes over a frictionless vertical pole. The mass of the ring is 0.100 kg. The spring may be assumed to be massless. Initially the ring is held horizontal so that the length PA is 0.300 m. The ring is then allowed to drop. Figure 5.26 For question 26. -2). (a) Calculate (using g = 10.0 ms (i) the speed of the ring when it reaches point B, a vertical distance of 0.400m from point A; (ii) the magnitude of the force exerted on the ring by the pole at point B; (iii) the acceleration (magnitude and direction) of the ring at B; (iv) the largest distance below A to which the ring falls (use your graphical calculator); (v) the maximum speed of the ring during its fall and the distance fallen when this happens (use your graphical calculator). (b) Using your graphical calculator, plot the speed of the ring as a function of distance fallen from A. (c) The ring will start moving upwards after reaching its lowest point. Discuss whether the ring will move higher than point A or whether it will stop at A. i ...
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This note was uploaded on 03/01/2011 for the course PHYSICS 121 taught by Professor Rothberg during the Spring '11 term at University of Washington.

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