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notes_7 - CONSERVATION OF MECHANICAL ENERGY Consider a...

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Unformatted text preview: CONSERVATION OF MECHANICAL ENERGY Consider a system consisting of an object and the Earth. We release the object from a height h above the ground. We take the ground as the zero of potential energy. CHAPTER 7 CONSERVATION OF ENERGY • Conservation of mechanical energy • Conservation of total energy of a system o various examples U1 = mgh : K1 = 0 v1 = 0 v 1 U = mgy : K = mv 2 2 v2 h 1 U2 = 0: K 2 = mv 2 2 2 y • Origin of friction 0 2 2 At the bottom: ( v = v o + 2a∆y ) ⇒ v 22 = 2 gh 1 ∴ K2 = mv 22 = mgh . 2 At any position y: v 2 = 2 g( h − y ) 1 ∴ K = mv 2 = mg( h − y ). 2 1 See how the individual parts of the mechanical energy vary with time as the objects falls ... v1 = 0 U1 = mgh : K1 = 0 m h h v 1 U = mgy : K = mv 2 = mg( h − y) 2 y v2 1 U2 = 0: K 2 = mv 2 2 = mgh 2 E mech = K + U h→ 1 K = mv 2 2 But we note that U = mgy K1 + U1 = K2 + U2 = K + U , t i.e., K + U = constant ⇒ E mech . We define E mech as the total mechanical energy of the system (i.e., the sum of the potential and kinetic energies). Since E mech = K + U ⇒ constant (in time) ∆E mech = ∆K + ∆U = 0 ∴ ∆K = − ∆ U . In this case: E mech = mgh , the initial energy of the i.e., the increase in kinetic energy equals the decrease in system. potential energy. Conservation of mechanical energy 2 Note that we can derive the conservation law in another way ... The work done by the Other examples of conservation of energy ... Simple pendulum gravitational force on the v1 = 0 h object is W1→2 = mgh . But, the change in potential v B A C h energy of the Earth-object system is v2 ∆U1→2 = U2 − U1 HMM12VD1.MOV Energy E mech = K + UG = mgh = 0 − mgh = − mgh , i.e., a decrease. UG ∴ W1→2 = − ∆U1→2. K Using the work-kinetic energy theorem we get W1→2 = ∆K1→2 = − ∆U1→2, i.e., the increase in kinetic energy from 1 → 2 is equal to the decrease in potential energy, as before. Note that mechanical energy (PE + KE) is conserved only if conservative forces are involved. t A → B → C → B → A E mech = K + UG ⇒ constant ( = mgh ) Continuous conversion of potential energy to kinetic energy and vice versa. 3 Oscillating spring When a mass (M) C is hung from a ys = kMg ys y=0 spring, the spring M extends to an y=0 yo oscillates up-and-down (with simple harmonic motion). D A equilibrium position, which we define at y = 0 . If the spring is further extended, to y = − y o , and released, it UE = 0 B A Energy 1 E total = kyo 2 2 E total UE C y=0 yo B UE = 0 K D A t A A There are two contributions to the mechanical energy, the kinetic energy of the mass, K = 12 Mv 2 , and the elastic energy associated with the spring, UE = 12 ky 2 , () () where k is the spring constant and y is the instantaneous position measured from the equilibrium position. B C D A B C D A The total energy is the initial energy, i.e., 1 E total = ky o2, 2 where y o is the initial displacement from the equilibrium position. Note that K + UE = constant = E total . 4 Conservation of mechanical energy gives us another way to determine velocities, displacements, etc. l T 30o l cos 30o h = l − l cos 30o v = 4.5m/s 2 1 mg mg Question 1: A pendulum consists of a 2kg mass attached to a string, of negligible mass, with a length of 3m. The The total mechanical energy at any point is: E = K + U. mass is struck horizontally so that it has an initial horizontal velocity of 4.5m/s. At the point where the Take the zero of potential energy at 1 , i.e., put U1 = 0. string makes an angle of 30o with the vertical, what is (a) Then the speed of the mass, (b) the potential energy of the at 1 ... 1 K1 = mv12 : U1 = 0 . 2 at 2 ... 1 K2 = mv 22 : U2 = mgh . 2 system, and (c) the tension in the string? (d) What is the maximum angle achieved by the string? But energy is conserved ... there are no other external forces, so E1 = E2. 5 T l T l 30o l cos 30o h = l − l cos 30 v = 4.5m/s 2 mg T 1 o mg (a) h = l − l cos 30o v = 4.5m/s 1 mg (c) Identify forces acting on the bob at 30o ∴ K1 + U1 = K2 + U2 1 1 i.e., mv12 = mv 22 + mgh . 2 2 1 1 ∴ × 2 × (4.5)2 = × 2 × v 22 2 2 +2 × 9.81 × 3(1 − cos 30o ), i.e., v 2 = 3.52 m/s. o (b) U2 = mgh = 2 × 9.81 × 3(1 − cos 30 ) = 7.89J . mg 2 mg 30o 30o l cos 30o T − mg cos 30o = mv 22 l ∴ T = 25.3N. (d) At maximum height: K = 0, so U = K1, 1 i.e., mgh max = mgl(1 − cos θ max ) = mv12. 2 v2 (4.5)2 ∴ cos θ max = 1 − 1 = 1 − = 0.656 , 2 gL 2 × 9.81 × 3 i.e., θ max = 49.0o. 6 k = 400N/m 5m x (a) At the top 5m k = 400N/m x Question 2: A 3kg object is released from rest at a height of 5m on a curved frictionless ramp, as shown above. At the foot of the ramp, there is a spring with a force constant of k = 400N/m. The object slides down the ramp into the spring, compressing it a distance x before coming momentarily to rest. (a) What is the distance x? (b) What happens to the object after it comes to rest? UG = 0 UG = mgh = 3 × 9.81 × 5 = 147.2J , K = 0 and UE = 0. ∴ E top = K + UG + UE = 147.2J. At the bottom, when the object is at rest against the spring, 1 UG = 0 , K = 0 and UE = kx 2. 2 ∴ E bottom = K + UG + UE = UE = 200 x 2J . But mechanical energy is conserved ... ∴ E bottom = E top , i.e., 200 x 2 = 147.2, ∴ x = 0.858 m. (b) The object will retrace its path back up to the start position at 5m (energy is conserved). Note: with friction on the track then mechanical energy will be “lost” (converted to heat), so x < 0.858 m, as the frictional force is non-conservative. 7 (a) Put the zero of gravitational potential energy at the ground. K = 0 : UE = 0 : UG = mgh = 78480J Question 3: A physics student, with mass 80kg, does a l = 50 m bungee jump from a platform 100m above the ground. If the rubber rope has an unstretched length of 50m and a h = 100m ∆l spring constant k = 200N/m, (a) how far above the ground will the student be at his lowest point? (b) How lowest point yo far above the ground will he achieve his greatest speed? (c) What is his greatest speed? 1 K = 0 : UE = k( ∆l) 2 : UG = mgyo 2 UG = 0 The total energy is 78480J, which is conserved throughout the jump. At his lowest point 1 K + k ( ∆l)2 + mgy o = 78480, 2 but K = 0 and ∆l = h − l − y o = 50 − y o. ∴100(50 − y o )2 + 80 × 9.81y o = 78480 , i.e., 100 y o2 − (10000 − 784.8) y o + (250000 − 78480) = 0. ∴ y o2 − 92.15y o + 1715.2 = 0 , i.e., y o = 92.15 ± (92.15)2 − 4 × 1715.2 . 2 8 The two solutions are: y o = 66.27 m and y o = 25.89 m. Clearly, the appropriate solution is y o = 25.89 m. Substituting for l′ = h − 50 − y = 50 − y , we find K + 100(50 − y )2 + 784.8 y = 78480 , i.e., K = 78480 − 100(50 − y )2 − 784.8 y We will return to the other solution in a moment. (b) To determine the position where the jumper’s speed = −171520 + 9215.2 y − 100 y 2 , so dK = 9215.2 − 200 y . dy is greatest, we will get an expression for the kinetic energy (K) in terms of the vertical position (y) and set dK = 0, i.e., where K is a maximum. dy K = 0 : UE = 0 : UG = mgh = 78480J l = 50m h = 100m max speed l′ y The extremum occurs when dK = 0, dy i.e., at y = 46.08 m. Check that this is a maximum: d 2K = −200 , i.e., < 0. dy 2 We will find the result by another method in a moment. 1 K : UE = kl ′ 2 : UG = mgy 2 UG = 0 Since energy is conserved 1 K + kl′2 + mgy = 78480. 2 (c) Since K max = 78480 − 100(50 − y )2 − 784.8 y when y = 46.08 m, we have 1 mv max2 = 40780J , 2 i.e., v max = 2 × 40780 = 31.9m/s. 80 9 DISCUSSION PROBLEM [7.1]: K = 0 : UE = 0 : UG = mgh = 78480J l = 50 m h = 100m DISCUSSION PROBLEM [7.2]: ∆l lowest point yo 1 K = 0 : UE = k( ∆l) 2 : UG = mgyo 2 UG = 0 Can you think of an alternative method for determining the position where the jumper’s speed is greatest? In part (a) of the previous problem we found two values for the lowest point y o, i.e., y o = 25.89 m and y o = 66.27 m. Clearly, the appropriate solution is the former. What does the latter solution correspond to? 10 When non-conservative forces are involved, the situation is a little more complicated: FTR is the result of a (downward) force F acting on the pedal(s) that is transmitted by a chain to a geared cog wheel attached to the axle of the wheel, which causes it to rotate. Case Study ... What makes a bicycle move when you pedal it? θ Motion F Force of tire on road FTR F Force of road on tire FRT By Newton’s 3rd Law FRT = FTR . If there is no slip (only static friction) the point of contact of tire does not move relative to the road. Therefore neither FTR nor FRT do any work! But it is FRT that “propels” the bike forward (⇒ ma ). If FRT does no work how can the kinetic energy change ... ∆K = W ? Let’s look at the origin of FTR . FTR FRT Since the force F - produced by muscle-power - moves through a displacement it does work. It is this work that propels the bicycle. Where does the force (work) originate? ... from energy supplied by the rider’s muscles (i.e., from food!) ... and it is a non-conservative force. It is the work done by the rider that results in the motion, i.e., Wrider ⇒ ∆K . So the rider has to be part of the system. 11 Thus, a more general statement of energy Let’s look at some examples: conservation must involve • with chemical changes the total energy of a system • with frictional forces (e.g., bike + Earth + rider ...) Esystem = ∑ Ei = E mech + Echem + K i If there is no input to the system from any other external sources (e.g., wind, a push, etc.), then Esystem = constant i.e., ∆Esystem = 0 . Question 4: A physics student, with mass 60kg, climbs a 120m high hill. (a) What is the increase in his gravitational potential energy? (b) Where does this energy come from? So an increase in one form of energy (e.g., mechanical) is (c) The student’s body is compensated by a decrease in another form (e.g., energy 20% efficient, i.e., for stored by rider. every 100J of chemical energy expended only 20J If an external source does work ( Wext ) on a system then: Wext = ∆Esystem . are converted to mechanical energy, with the remainder (80J) resulting in This is the more general form of the work-energy thermal energy. How much chemical energy is theorem. An example might be the action of a wind expended by the student during the climb? acting on the rider. 12 (a) ∆U = U2 − U1 = mgh = 80 × 9.81 × 1200 = 94.2 kJ . The total mechanical energy change ⇒ 94.2 kJ . DISCUSSION PROBLEM [7.3]: (b) The system is the climber and the Earth. With no external forces Esys = E mech + Eclimber , where Eclimber is the energy stored within the climber. Since ∆Esys = 0, ∆E mech = − ∆Eclimber , i.e., the energy comes from a decrease in (chemical) energy stored by the climber, which he got from metabolizing food. (c) 100J (Echem) ⇒ 20J ( E mech ) + 80J ( E therm) ∴ 1J ( E mech) requires 5J ( Echem) Why does a cyclist need to keep pedaling when traveling on a level road, even though she’s not changing her speed (i.e., with no change in ∆K )? Isn’t the workenergy theorem ( W ⇒ ∆K ) violated? Explain. i.e., 94.2kJ ( E mech) requires 5 × 94.2 = 471kJ of chemical energy. 13 N θ mg 0.50m 2.0 m 0.50 m θ 2.0m Question 5: A physics student has been given the task of designing a system to deliver a glass of beer along a bar to a designated spot. The beer mug, of mass 1.5kg is released from the top of a ramp, which is 0.50m above There are two contributions to the total energy of the system; E mech and E therm (assuming that the work done against friction produces heat. ∴ Esys = E mech + E therm . If there are no external forces, then Esys is constant, i.e., ∆Esys = ∆E mech + ∆E therm = 0. where it should come to a stop 2.0m from the bottom of Now, ∆E mech = ∆K + ∆U and ∆E therm = f k .∆s, where f k is the frictional force and ∆s is the distance through the ramp. If the coefficient of kinetic friction between which the frictional force is active. the bar. The mug slides down the ramp and onto the bar, the mug and the surfaces of the ramp and bar in 0.15, (a) what is the required value of the angle θ ? (b) What is the speed of the mug at the bottom of the ramp? (a) In this problem, ∆E therm comprises two components: (1) down the ramp and (2) along the bar. 0.50 (1) down the ramp: ∆s1 = and sin θ f k1 = µ mN = µ k mg cos θ . 14 N θ mg 0.50m 2.0m N 0.50 0.15 × 1.5 × 9.81 × 0.5 ∴ ∆E therm1 = µ k mg cos θ. = sin θ tan θ 1.104 = J. tan θ (2) Along the bar: ∆s2 = 2.0 m and f k2 = µ k mg ∴ ∆E therm2 = µ k mg × 2.0 = 0.15 × 1.5 × 9.81 × 2.0 = 4.415J. 1.104 ∴ ∆E therm = + 4.415J . tan θ Using conservation of energy ∆K + ∆Ug + ∆E therm = 0, but ∆K = 0 , so ∆E therm = − ∆Ug. ∆Ug = − mgh = −1.5 × 9.81 × 0.5 = −7.358J , θ mg 0.50m 2.0m (b) On the ramp: ∆Esys = ∆E mech + ∆E therm1 = 0, i.e., ∆K1 + ∆Ug + ∆E therm1 = 0 . 1 1.104 ∴ ∆K1 = mv 2 = − ∆Ug − ∆E therm1 = mgh − 2 tan θ = 7.385 − 2.937 = 4.448J. 2 × 4.448 ∴ v2 = = 5.93 ⇒ v = 2.44 m/s. 1.50 1.104 + 4.415 = 7.385, tan θ 1.104 = 20.6o . i.e., θ = tan −1 7.358 − 4.415 15 Hey .. before going any further ... what’s the origin of friction? DISCUSSION PROBLEM [7.4]: Magnum XL-200, Cedar Point Park, Ohio Friction results from attractive forces between atoms on the two surfaces. When sliding occurs work is done to • What forces are involved in a roller coaster? stretch the “bonds”; they break and the participating • Are the forces conservative or non-conservative ? atoms vibrate. ~ Then why do the surfaces get hot ? ~ ∆K ⇒ ∆ T (kinetic theory). • What changes of energy are involved? Increased vibration (velocity) ⇒ increased temperature. 16 1 start position 5.0m ∆y 2 max kinetic energy yo 3 max compression (a) Let Ug = 0 at position 2 . Then conservation of Question 6: A block of mass 2.4kg is dropped onto a energy gives: spring, with a spring constant of 400N/m, from a height K1 + Ug1 + Ue1 = K2 + Ug2 + Ue2, of 5.0m above the top of the spring. (a) What is the 1 i.e., mg(5 + ∆y ) = K2 + k ( ∆y )2 . 2 1 ∴ K2 = mg(5 + ∆y ) − k ( ∆y )2 . 2 maximum kinetic energy of the block? (b) What is the maximum compression of the spring? This is an interesting problem as we will use different positions for the zero of gravitational potential energy in parts (a) and (b). For maximum kinetic energy: dK2 = 0, d ( ∆y ) i.e., mg − k∆y = 0 ⇒ ∆y = mg 2.4 × 9.81 = k 400 = 0.059 m. 17 1 But K2 = mg(5 + ∆y ) − k ( ∆y )2 2 1 = 2.4 × 9.81 × 5.059 − × 400 × (0.059)2 2 = 118.4J . ( v max = 9.93m/s.) (b) Now we let Ug = 0 at position 3 . i.e., y o = 23.54 ± (23.54)2 + 4 × 200 × 117.7 2 × 200 ∴ y o = 0.828 m or y o = −0.711m. Clearly, the former is the physically meaningful root and represents the maximum compression of the spring. 1 start position 5.0 m ∆y yo If the mass were simply placed on the top of the spring, the equilibrium position would occur when the spring is 2 max kinetic energy 3 max compression compressed a distance ∆l (from the top of the spring), so the net force acting on the mass is zero, i.e., when k∆l = mg . Then conservation of energy gives: K1 + Ug1 + Ue1 = K3 + Ug3 + Ue3, 1 i.e., mg(5 + y o ) = ky o2. 2 ∴ ∆l = mg 2.4 × 9.81 = = 0.059 m. 400 k Note that this is the position where the kinetic energy was a maximum in part (a). ∴ 200 y o2 − 23.54 y o − 117.7 = 0 18 ...
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This note was uploaded on 07/13/2011 for the course PHY 2048 taught by Professor Guzman during the Spring '08 term at FAU.

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