Energy Power Lana heridan De Anza College Nov 1, 2017
Overview Power Practice problems!
Power Power the rate of energy transfer to a system.
Power Power the rate of energy transfer to a system. The instantaneous power is defined as P = de dt
Power Power the rate of energy transfer to a system. The instantaneous power is defined as P = de dt The average power is P = E t Units?
Power Power the rate of energy transfer to a system. The instantaneous power is defined as P = de dt The average power is P = E t Units? The Watt. 1 J/s = 1 W
Power Most often we are interested in the rate of work done on a system P = dw dt From the definition of work W = F dr: P = dw dt = F dr dt This gives another expression for power P = F v
39. A 3.50-kN piano is lifted by three workers at constant speed to an apartment 25.0 m above the street using a pulley system fastened to the roof of the building. Each worker is able to deliver 165 W of power, and the pulley system is 75.0% efficient (so that 25.0% of the mechanical energy is transformed to other forms due to fricer n- Power ns -m his he ass er m s it wn ine ns air is he se- cise physiology, power is often measured in kcal/h rather than in watts. Here 1 kcal 5 1 nutritionist s Calorie = 4 186 J.) Walking at 3.00 mi/h requires about 220 kcal/h. It is interesting to compare these values with the energy consumption required for travel by car. Gasoline yields about 1.30 3 10 8 J/gal. Find the fuel economy in equivalent miles per gallon for a person Page 239, #38 (a) walking and (b) bicycling. 38. A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of 1.75 m/s. (a) What is the average power of the elevator motor during this time interval? (b) How does this power compare with the motor power when the elevator moves at its cruising speed?
Power Average power, P avg = W t.
Power Average power, P avg = W t. What is the work done by the motor lifting the elevator? W = F T dr
Power Average power, P avg = W t. What is the work done by the motor lifting the elevator? W = F T dr = (mg + ma)dy
Power Average power, P avg = W t. What is the work done by the motor lifting the elevator? W = F T dr = (mg + ma)dy = mgdy + m dv dt dy = mg y + 1 2 m(v 2 f v 2 i ) (W = U g + K)
Power Average power, P avg = W t. What is the work done by the motor lifting the elevator? W = F T dr = (mg + ma)dy = mgdy + m dv dt dy = mg y + 1 2 m(v 2 f v 2 i ) (W = U g + K) v i = 0, v f = 1.75 m/s, so y = v avg t = 0+1.75 2 (3) = 2.625 m.
Power W = mg y + 1 2 m(v 2 f v 2 i ) = 17716 J P = W = 17716 J t 3 s = 5.91 10 3 W = 5.91 kw
Power W = mg y + 1 2 m(v 2 f v 2 i ) = 17716 J P = W = 17716 J t 3 s = 5.91 10 3 W = 5.91 kw What about moving upward at constant speed?
Power What about moving upward at constant speed?
Power What about moving upward at constant speed? F = mg j, v = 1.75 j m/s P = F v = (mg)v = 1.11 10 4 W = 11.1 kw
Conservation of Energy That pretty much concludes everything covered in chapter 8. For the rest of the lecture we will look at applying ideas already introduced.
w much e skatey in the at point Caution: ired skill mum. (b) In the What If? section of this example, we explored the effects of an increased friction force of 10.0 N. At what position of the block does its maximum speed occur in this situation? Drag and Power (page 241) le of old motion by the 1 x2 5 54. As it plows a parking lot, a Q/C snowplow pushes an evergrowing pile of snow in front of it. uppose a car moving through the air is similarly modeled as a cylinder of area A pushing a growing disk of air in front of it. The originally stationary air is set into motion at the constant speed v of the cylinder as shown in Figure P8.54. In a time interval Dt, a new disk of air of mass Dm must be moved a distance v Dt and hence must be given a kinetic energy 1 2 1Dm2v2. Using this model, show that the car s power loss owing to air resistance is 1 2rAv 3 and that the resistive force acting on the car is 1 2rAv 2, where r is the density of air. Compare this result with the empirical expression 1 2DrAv 2 for the resistive force. 55. A wind turbine on a wind farm turns in response to a force of high-speed air resistance, R 5 1 DrAv 2. The A v t Figure P8.54 v
Drag and Power What is the volume of air being accelerated in time t? V = Av t The density of air, ρ = m V, so m = ρav t Then, K = 1 2 ( m)v 2 = 1 2 ρav 3 t Power is the rate of transfer of energy. The car is losing energy K in time t, so P = dk dt = 1 2 ρav 3
Drag and Power Force? P = F v 1 2 ρav 3 = Fv Giving a force, F = 1 2 ρav 2 c.f. the Drag Equation: F = 1 2 DρAv 2. The only missing piece is the drag coefficient. This model is not refined enough to deal with different shapes of object, so that s not really surprising. (Basically, we have D = 1, about right for a flat-fronted object.)
a nonconservative force acts. Example: Block pulled across surface g along a freeway at 65 mi/h. Your car has kinetic stop Example because of 8.4, congestion Page 224 in traffic. Where is ce had? (a) It is all in internal energy in the road. in the Atires. 6.0 kg(c) block ome initially of it at has rest transformed is pulled to the to right along a t transferred horizontal away surface by mechanical by a constantwaves. horizontal (d) force It is all of 12 N. r by various mechanisms. Find the speed of the block after it has moved 3.0 m if the surfaces in contact have a coefficient of kinetic friction of µ k = 0.15. AM n v f ntal surface by a f k F faces in contact a mg x
Example 8.4
surface Example by a 8.4 f k F es in contact x uppose the force F is applied mg at an angle θ. At what angle should the force be applied to achieve the largest possible speed after the a block has moved 3.0 m to the right? Example 8.4) lled to the right face by a conal force. (b) The s at an angle u tal. b f k n mg F u x v f
Example 8.4
Example 8.4
is k 5 50 N/m as shown in Figure 8.11. sion AM Example 8.8: pring Collisions and Friction calculate the maximum compression of the spring after the collision. initial Avelocity block having v 5 1.2 a m/s massto of the 0.80 right kgand is given collides an with initial a spring velocity whose is k 5 50 N/m as shown in Figure 8.11. x 0 v = 1.2 m/s to the right, just as it collides with a spring whose re 8.11 (Example v, calculate massthe is maximum negligible compression and whose force of the constant spring after is k the = 50 collision. N/m. A block sliding on a 1 E mv 2 ionless, horizontal a 2 ce collides with a x 0 re spring. 8.11 (a)(example Initially, v v A echanical block sliding energy on a is 1 ionless, horizontal a E mv 2 1 inetic energy. (b) The kx 2 b 2 2 ce anical collides energy with is a the of spring. the kinetic (a) Initially, energy x e v echanical block and energy the elasotential 1 is E mv 2 1 inetic energy. energy (b) in The the v kx 2 b 0 2 2 hanical g. (c) The energy energy is the is 1 E kx 2 ely of the potential kinetic energy. c 2 max x e he block energy and is the transed back energy to the kinetic the elasotential x max v v gy 0 g. of (c) the The block. energy is v 1 E 2 kx 2 total ely potential energy of energy. the c max 1 E mv he energy is transed back the to the motion. k = 0.50. What is the maximum compression x in the x 2 1 2 m remains constant d mv 2 2 surface, µ ughout kinetic spring? gy of the block. v v total, the energy total mechanical of the energy of the system before the collision is just 1 mv 2. A constant force of kinetic friction acts between the block and the
Example 8.8
Example 8.7: Box liding Down an Incline A 3.00 kg crate slides down a ramp. The ramp is 1.00 m in length and inclined at an angle of 30.0. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it AMleaves the ramp. 0 m in length and he crate starts from f magnitude 5.00 N, zontal floor after it v i 0.500 m 0 d 1.00 m 30.0 v f e crate at the bot- Use energy methods Figure to8.10 determine (Example the 8.7) A speed crate slides of the crate at the amp in Figure 8.10. down a ramp under the influence of gravity. will slide. bottom of the ramp. The potential energy of the system decreases, whereas the kinetic energy increases. Earth as an isolated
Example 8.7: Box liding Down an Incline
Example 8.7: Part 2 A 3.00 kg crate slides down a ramp. The ramp is 1.00 m in length and inclined at an angle of 30.0. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it AMleaves the ramp. 0 m in length and he crate starts from f magnitude 5.00 N, zontal floor after it v i 0.500 m 0 d 1.00 m 30.0 v f e crate at the bot- How far does the Figure crate8.10 slide (Example on the8.7) horizontal A crate slides floor if it continues amp in Figure 8.10. down a ramp under the influence of gravity. will slide. to experience a The friction potential force energy of magnitude of the system decreases, 5.00 N? whereas the kinetic energy increases. Earth as an isolated
Example 8.7
30.0 cm (Fig. P8.43). Calculate (a) the gravitational Mass potential in a hemisphere energy of the block Earth system when the block is at point relative to point, (b) the kinetic energy of the block at point, (c) its speed at point, and (d) its kinetic energy and the potential energy Page when 240, the #43 block and is at #44 point. R 2R/3 Figure P8.43 Problems 43 and 44. 44. What If? The block of mass m 5 200 g described in Q/C Problem 43 (Fig. P8.43) is released from rest at point, and the surface of the bowl is rough. The block s
Additional Problems Mass in a hemisphere (discuss, don t calculate!) 42. Make an order-of-magnitude estimate of your power BIO output as you climb stairs. In your solution, state the physical quantities you take as data and the values you measure or estimate for them. Do you consider your peak power or your sustainable power? Page 240, #43 43. A small block of mass m 5 200 g is released from rest at point along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R 5 30.0 cm (Fig. P8.43). Calculate (a) the gravitational potential energy of the block Earth system when the block is at point relative to point, (b) the kinetic energy of the block at point, (c) its speed at point, and (d) its kinetic energy and the potential energy when the block is at point. R 2R/3 at ta ab ha ab ed to izo in flo te th (c) do of en (e 47. A M va is en
Mass in a hemisphere Page 240, #44 R 2R/3 Figure P8.43 Problems 43 and 44. 44. What If? The block of mass m 5 200 g described in Q/C Problem 43 (Fig. P8.43) is released from rest at point, and the surface of the bowl is rough. The block s speed at point is 1.50 m/s. (a) What is its kinetic energy at point? (b) How much mechanical energy is transformed into internal energy as the block moves from point to point? (c) Is it possible to determine the coefficient of friction from these results in any simple manner? (d) Explain your answer to part (c). M va is en ti (c an t 48. W h re ro th h m b m
Mass in a hemisphere
ummary Power Energy problem solving Next Test Friday, Nov 3, Chapters 6-8, and pulleys and friction from Ch 5. (Uncollected) Homework erway & Jewett, Read Chapter 8, understand the examples. Go back over Chapters 6-8 and identify anything unclear. Ch 8, onward from page 236. Probs: 43, 57, 61, 65, 67