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Energy Topics: • Important forms of energy • How energy can be transformed and transferred • Definition of work Definition of work • Concepts of kinetic, potential, and thermal energy • The law of conservation of energy • Elastic collisions Sample question: When flexible poles became available for pole vaulting, athletes were able to clear much higher bars. How can we explain this using energy concepts? A “Natural Money” Called Energy Income System Liquid Asset: Cash Saved Asset: Stocks Transfers into and out of system Transformations Transformations within system Key concepts: • Definition of the system. Expenses • Transformations within the system. y • Transfers between the system and the environment. Forms of Energy Mechanical Energy gy Ug K Thermal E Energy Us Other forms include Eth Echem Enuclear The Basic Energy Model The unit of work and energy is the joule (J). 1 J = 1 Nm = 1 kg m2/s2. Checking Understanding A skier is moving down a slope at a constant speed speed. What energy transformation is taking place? A. K → U g B. U g → Eth C. U s → U g D. Ug → K E. K → Eth Answer A skier is moving down a slope at a constant speed speed. What energy transformation is taking place? B. U g → Eth Slide 10-13 Checking Understanding A child is on a playground swing swing, motionless at the highest point of his arc. As he swings back down to the lowest point of his motion, what energy transformation is taking place? A. B. C C. D. E. K → Ug U g → Eth Us → Ug Ug → K K → Eth Slide 10-14 Answer A child is on a playground swing swing, motionless at the highest point of his arc. As he swings back down to the lowest point of his motion, what energy transformation is taking place? D. U g → K Slide 10-15 Kinetic Energy Kinetic Energy 1 2 K = mv 2 is an object’s translational kinetic energy. This is the energy an object has because of its state of motion. It can be shown that, in general Wnet = ΔK . Work-Kinetic Energy Theorem We have seen earlier that K f − K i = Wnet . We define the change in kinetic energy as ΔK = K f − K i . The equation above becomes the work-kinetic k ki i energy theorem h : ΔK = K f − K i = Wnet ⎡Change in the kinetic ⎤ ⎡ net work done on ⎤ ⎢energy of a particle ⎥ = ⎢ the particle ⎥ ⎣ ⎦ ⎣ ⎦ The work-kinetic energy theorem holds for both positive and negative values of Wnet . If Wnet > 0 → K f − K i > 0 → K f > K i If Wnet < 0 → K f − K i < 0 → K f < K i Example: The extinction of the dinosaurs and the majority of species on Earth in the Cretaceous Period (65 Myr ago) is thought to have been caused by an asteroid striking the Earth near the Yucatan Peninsula The resulting ejecta caused widespread global climate Earth near the Yucatan Peninsula. The resulting ejecta caused widespread global climate change. If the mass of the asteroid was 1016 kg (diameter in the range of 4‐9 miles) and had a speed of 30.0 km/sec, what was the asteroid’s kinetic energy? ( )( 1 2 1 16 K = mv = 10 kg k 30 ×103 m/s / 2 2 = 4.5 × 10 24 J ) 2 This is equivalent to ~109 Megatons of TNT. Work by a Constant Force Work by a Constant Force Work is an energy transfer by the application of a force. For work to be done there must be a nonzero displacement. The unit of work and energy is the joule (J). 1 J = 1 Nm = 1 kg m2/s2. 12 m m Finding an Expression for Work : Consider a bead of mass m that can move without friction along a straight wire along G the x-axis. A constant force F applied at an angle φ to the wire is acting on the bead. W = Fd cos ϕ G G W = F ⋅d Example: A ball is tossed straight up. What is the work done by the force of gravity on the ball as it rises? y Δy FBD for rising ball: x w Wg = wΔy cos180° = −mgΔy Example: A box of mass m is towed up a frictionless incline at constant speed. The applied force F is parallel to the incline. What is the net work done on the box? y F An FBD for the box: N F x θ θ w Apply Newton’s 2nd Law: ∑F ∑F x = F − w sin θ = 0 y = N − w cos θ = 0 Work by a Variable Force Work by a Variable Force Work can be calculated by finding the area underneath a plot of the applied force in the Work can be calculated by finding the area underneath a plot of the applied force in the direction of the displacement versus the displacement. 16 Example: What is the work done by the variable force shown below? Fx (N) F3 F2 F1 x1 x2 x3 x (m) x (m) The work done by F1 is W1 = F1 ( x1 − 0 ) The work done by F2 is W2 = F2 ( x2 − x1 ) The work done by F3 is W3 = F3 ( x3 − x2 ) The net work is then W1+W2+W3. 17 Gravitational Potential Energy Gravitational Potential Energy Objects have potential energy because of their location (or configuration). There are potential energies associated with different (but not all!) forces. Such a force is called a conservative force. In general Wcons = − ΔU The change in gravitational potential energy (only near the surface of the Earth) is ΔU g = mgΔy where Δy is the change in the object’s vertical position with respect to some reference point that you are free to choose. 19 Example: What is the change in gravitational potential energy of the box if it is placed on the table? The table is 1.0 m tall and the mass of the box is 1.0 kg. First: Choose the reference level at the floor. U = 0 here. ΔU g = mgΔy = mg ( y f − yi ) ( ) = (1.0 kg ) 9.8 m/s 2 (1.0 m − 0 m ) = +9.8 J 20 Example continued: Now take the reference level (U = 0) to be on top of the table so that yi = −1.0 m and yf = 0.0 m. ΔU g = mgΔy = mg ( y f − yi ) ( ) = (1 kg ) 9.8 m/s 2 (0.0m − (− 1.0 m )) = +9.8 J The results do not depend on the location of The results do not depend on the location of U = 0. Mechanical energy is E = K +U Whenever nonconservative forces do no work, the mechanical energy of a system is conserved. That is Ei = Ef or ΔK = −ΔU. 22 Example: A cart starts from position 4 with v = 15.0 m/s to the left. Find the speed of the cart at positions 1, 2, and 3. Ignore friction. p , , g E4 = E3 U 4 + K 4 = U 3 + K3 1 2 1 2 mgy4 + mv4 = mgy3 + mv3 2 2 v3 = v42 + 2 g ( y4 − y3 ) = 20.5 m/s / Example continued: E4 = E2 Or use E3=E2 U 4 + K4 = U 2 + K2 1 1 mgy4 + mv42 = mgy2 + mv22 2 2 v2 = v42 + 2 g ( y4 − y2 ) = 18.0 m/s E4 = E1 U 4 + K 4 = U 1 + K1 1 2 1 2 mgy4 + mv4 = mgy1 + mv1 2 2 v1 = v42 + 2 g ( y4 − y1 ) = 24.8 m/s / Or use E3=E1 E2=E1 Elastic potential energy / Spring force By hanging masses on a spring we find that stretch ∝ applied force. This is Hooke’s law. li d f Thi i H k ’ l For an ideal spring: F d l k x = ‐kx Fx is the magnitude of the force exerted by the free end of the is the magnitude of the force exerted by the free end of the spring, x is the measured stretch of the spring, and k is the spring constant (a constant of proportionality; its units are N/m). A larger value of k implies a stiffer spring A larger value of k implies a stiffer spring. 25 Example: (a) If forces of 5.0 N applied to each end of a spring cause the spring to stretch 3.5 cm from its relaxed length, how far does a force of 7.0 N cause the same spring to 3.5 cm from its relaxed length, how far does a force of 7.0 N cause the same spring to stretch? For springs F∝x This allows us to write For springs F∝x. This allows us to write Solving for x2: F1 x1 = . F2 x2 F2 ⎛ 7.0 N ⎞ x1 = ⎜ x2 = ⎟(3.5 cm ) = 4.9 cm. F1 ⎝ 5.0 N ⎠ Example continued: (b) What is the spring constant of this spring? F1 5.0 N k= = = 1.43 N/cm. x1 3.5 cm Or F2 7.0 N k= = = 1.43 N/cm. x2 4.9 cm Example: An ideal spring has k = 20.0 N/m. What is the amount of work done (by an external agent) to stretch the spring 0.40 m from its relaxed length? g ) p g g Fx (N) kx1 x1=0.4 m =0 4 m x (m) W = Area under curve 1 1 2 1 2 = (kx1 )( x1 ) = kx1 = (20.0 N/m )(0.4 m ) = 1.6 J 2 2 2 28 Elastic potential energy Elastic potential energy The work done in stretching/compressing a spring transfers energy to the spring. 1 2 U s = kx 2 29 Example: A box of mass 0.25 kg slides along a horizontal, frictionless surface with a speed of 3.0 m/s. The box encounters a spring with k = 200 N/m. How far is the spring speed of 3.0 m/s. The box encounters a spring with k 200 N/m. How far is the spring compressed when the box is brought to rest? Ei = E f U i + Ki = U f + K f 1 2 1 2 0 + mv = kx + 0 2 2 ⎛ m⎞ ⎟v = 0.11 m x = ⎜⎜ ⎟ k ⎠ ⎝ 30 Conceptual Example A car sits at rest at the top of a hill hill. A small push sends it rolling down a hill. After its height has dropped by 5.0 m, it is moving at a good clip. Write down the equation for conservation ti off energy, noting ti the th choice h i off system, t the th initial i iti l and final states, and what energy transformation has taken place. Slide 10-17 Checking Understanding Each of the boxes, boxes with masses noted noted, is pulled for 10 m across a level, frictionless floor by the noted force. Which box experiences the largest change in kinetic energy? Slide 10-18 Answer Each of the boxes, boxes with masses noted noted, is pulled for 10 m across a level, frictionless floor by the noted force. Which box experiences the largest change in kinetic energy? Slide 10-19 Checking Understanding Each of the boxes, boxes with masses noted noted, is pulled for 10 m across a level, frictionless floor by the noted force. Which box experiences the smallest change in kinetic energy? Slide 10-20 Answer Each of the boxes, boxes with masses noted noted, is pulled for 10 m across a level, frictionless floor by the noted force. Which box experiences the smallest change in kinetic energy? Slide 10-21 Example A 200 g block on a frictionless surface is pushed against a spring with spring constant 500 N/m, compressing the spring by 2.0 cm. When the block is released, at what speed does it shoot h t away from f the th spring? i ? Slide 10-23 Example A 2.0 2 0 g desert locust can achieve a takeoff speed of 3.6 m/s (comparable to the best human jumpers) by using energy stored in an internal “spring” near the knee joint joint. A. When the locust jumps, what energy transformation takes place? B What B. Wh t is i the th minimum i i amountt off energy stored in the internal spring? C. If the locust were to make a vertical leap, h how hi high h could ld it jjump? ? IIgnore air i resistance and use conservation of energy concepts to solve this problem. D. If 50% of the initial kinetic energy is transformed to thermal energy because of air resistance, how high will the locust jump? Slide 10-24 Power • • Same mass... mass Both reach 60 mph... Same final kinetic energy, but different times mean different powers powers. Average Power Average Power ΔE Pav = Δt Instantaneous Power P = Fv cosθ Slide 10-25 Example : A race car with a mass of 500.0 kg completes a quarter‐mile (402 m) race in a time of 4 2 s starting from rest The car’ss final speed is 125 m/s. What is the engine time of 4.2 s starting from rest. The car final speed is 125 m/s What is the engine’ss average power output? Neglect friction and air resistance. ΔE ΔU + ΔK Pav = = Δt Δt 1 2 mv f ΔK 2 = = = 9.3 × 105 watts Δt Δt Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power? Car Mass (g) Speed (m/s) Time (s) A 100 3 2 B 200 2 2 C 200 2 3 D 300 2 3 E 300 1 4 Slide 10-26 Answer Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power? Car A Mass (g) Speed (m/s) Time (s) 100 3 2 Slide 10-27 Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the smallest power? Car Mass (g) Speed (m/s) Time (s) A 100 3 2 B 200 2 2 C 200 2 3 D 300 2 3 E 300 1 4 Slide 10-28 Answer Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the smallest power? Car E Mass (g) Speed (m/s) Time (s) 300 1 4 Slide 10-29 A typical human head has a mass of 5 5.0 0 kg kg. If the head is moving at some speed and strikes a fixed surface, it will come to rest. A helmet can help protect against injury; the foam in the helmet allows the head to come to rest over a longer distance distance, reducing the force on the head. The foam in helmets is generally designed to fail at a certain large force below the threshold of damage to th h the head. d If thi this fforce is i exceeded, d d th the ffoam b begins i tto compress. If the foam in a helmet compresses by 1.5 cm under a force of 2500 500 N (be (below o tthe e tthreshold es o d for o da damage age to tthe e head), ead), what at is s tthe e maximum speed the head could have on impact? Use energy concepts to solve this problem. Slide 10-31 Data D t for f one stage t off the th 2004 Tour T de d France F show h that th t Lance L Armstrong’s average speed was 15 m/s, and that keeping Lance and his bike moving at this zippy pace required a power of 450 W. A. What was the average forward force keeping Lance and his bike moving forward? B. To put this in perspective, compute what mass would have this weight. Slide 10-32 Question Each of the 1 1.0 0 kg boxes starts at rest and is then pushed for 2 2.0 0 m across a level, frictionless floor by a rope with the noted force. Which box has the highest final speed? Slide 10-33 Answer Each of the 1 1.0 0 kg boxes starts at rest and is then pushed for 2 2.0 0 m across a level, frictionless floor by a rope with the noted force. Which box has the highest final speed? Slide 10-34 Additional Clicker Questions Three balls are thrown off a cliff with the same speed speed, but in different directions. Which ball has the greatest speed just before it hits the ground? A. Ball A B Ball B. B ll B C. Ball C D. All balls have the same speed Slide 10-41 Answer Three balls are thrown off a cliff with the same speed speed, but in different directions. Which ball has the greatest speed just before it hits the ground? D. All balls have the same speed Slide 10-42 Questions A 20-cm-long 20 cm long spring is attached to a wall wall. When pulled horizontally with a force of 100 N, the spring stretches to a length of 22 cm. What is the value of the spring constant? A. 5000 N/m B. 500 N/m C. 454 N/m Slide 10-43 Answer A 20-cm-long 20 cm long spring is attached to a wall wall. When pulled horizontally with a force of 100 N, the spring stretches to a length of 22 cm. What is the value of the spring constant? C. 454 N/m Slide 10-44