Energy Practice. Lana Sheridan. Nov 2, De Anza College

Transcription

1 Energy Practice Lana heridan De Anza College Nov 2, 2017

2 Overview Practice problems!

3 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 = AM n v f ntal surface by a f k F faces in contact a mg x

4 Example 8.4 W = K + E int We can relate the final kinetic energy (starting from rest) to the forces: K 0 = W E int K = F dr f k s = F dr f k s 1 2 mv 2 = Fs µ k (mg)s 2(Fs µk (mg)s) v = m 2 ( (12 N)(3 m) (0.15)(6 kg)g(3 m) ) = = 1.8 m s 1 6 kg

5 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

6 Example 8.4 θ for largest v? W = K + E int Again, we d like an expression for K:

7 Example 8.4 θ for largest v? W = K + E int Again, we d like an expression for K: K = W E int = F dr f k s = Fs cos θ µ k (mg F sin θ)s where we noticed n = mg F sin θ.

8 Example 8.4 K = Fs cos θ µ k (mg F sin θ)s Maximize v, and therefore K, with respect to θ. Find the derivative, set it to zero:

9 Example 8.4 K = Fs cos θ µ k (mg F sin θ)s Maximize v, and therefore K, with respect to θ. Find the derivative, set it to zero: dk dθ = Fs sin θ µ k ( F cos θ)s = Fs(µ k cos θ sin θ)

10 Example 8.4 K = Fs cos θ µ k (mg F sin θ)s Maximize v, and therefore K, with respect to θ. Find the derivative, set it to zero: dk dθ = Fs sin θ µ k ( F cos θ)s = Fs(µ k cos θ sin θ) Fs(µ k cos θ sin θ) = 0 µ k cos θ = sin θ θ = tan 1 µ k = 8.5

11 pring and Friction Problem Page 240, Chapter 8, #21 A toy cannon uses a spring to project a 5.30-g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m. When the cannon is fired, the ball moves 15.0 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of N on the ball. (a) With what speed does the projectile leave the barrel of the cannon? (b) At what point does the ball have maximum speed? (c) What is this maximum speed?

12 pring and Friction Problem Page 240, Chapter 8, #21 A toy cannon uses a spring to project a 5.30-g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m. When the cannon is fired, the ball moves 15.0 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of N on the ball. (a) With what speed does the projectile leave the barrel of the cannon? (b) At what point does the ball have maximum speed? (c) What is this maximum speed? (a) v = 1.40 m/s

13 pring and Friction Problem Page 240, Chapter 8, #21 A toy cannon uses a spring to project a 5.30-g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m. When the cannon is fired, the ball moves 15.0 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of N on the ball. (a) With what speed does the projectile leave the barrel of the cannon? (b) At what point does the ball have maximum speed? (c) What is this maximum speed? (a) v = 1.40 m/s (b) s = 4.60 cm

14 pring and Friction Problem Page 240, Chapter 8, #21 A toy cannon uses a spring to project a 5.30-g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m. When the cannon is fired, the ball moves 15.0 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of N on the ball. (a) With what speed does the projectile leave the barrel of the cannon? (b) At what point does the ball have maximum speed? (c) What is this maximum speed? (a) v = 1.40 m/s (b) s = 4.60 cm (c) v = 1.79 m/s

15 is k 5 50 N/m as shown in Figure sion AM Example 8.8: pring Collisions and Friction calculate the maximum compression of the spring after the collision. initial Avelocity block having v 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 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 (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 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 = What is the maximum compression x in the x 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

16 Example 8.8 K + U + E int = 0

17 Example 8.8 K + U + E int = 0 (0 1 2 mv 2 i ) + ( 1 2 kx 2 f 0) + f k x f = kx 2 f + µ k mgx 1 2 mv 2 i = 0

18 Example 8.8 K + U + E int = 0 (0 1 2 mv 2 i ) + ( 1 2 kx 2 f 0) + f k x f = kx 2 f + µ k mgx 1 2 mv 2 i = 0 Quadratic expression. olution: x f = m = 9.2 cm (The other solution is x f = 0.25 m, what does that correspond to?)

19 3.20 m 2.00 m Isolated and Nonisolated system example Page 237, #7 Figure P Two objects are connected M by a light string passing over a light, frictionless pulley as shown in Figure P8.7. The object of mass m kg is released from rest at a height h m above the table. Using the isolated system model, (a) determine the speed of the object of mass m kg just as the 5.00-kg object hits the table and (b) find the maximum height above the table to which the 3.00-kg object rises. 8. Two objects are connected by a light string passing over a light, frictionless pulley as shown in Figure P8.7. The object of mass m 1 is released from rest at height m 2 m 1 h Figure P8.7 Problems 7 and 8. blocks block B leys. T rest so height are the block separa ection A sled kick im The co is the sle 13. A sled kick im ficient Use en moves 14. A crate M an init paralle

20 3.20 m 2.00 m Isolated and Nonisolated system example Page 237, #7 Figure P Two objects are connected M by a light string passing over a light, frictionless pulley as shown in Figure P8.7. The object of mass m kg is released from rest at a height h m above the table. Using the isolated system model, (a) determine the speed of the object of mass m kg just as the 5.00-kg object hits the table and (b) find the maximum height above the table to which the 3.00-kg object rises. 8. Two objects are connected by a light string passing answers: (a) over 4.43 a light, m/s frictionless (b) 5 m pulley as shown in Figure P8.7. The object of mass m 1 is released from rest at height m 2 m 1 h Figure P8.7 Problems 7 and 8. blocks block B leys. T rest so height are the block separa ection A sled kick im The co is the sle 13. A sled kick im ficient Use en moves 14. A crate M an init paralle

21 e Example: spring. To Energy what maximum with Circular height Motion release does it rise? all is Page fired 236, from Chapter a cannon 8, #5with muz- /s at an angle of with the horall is (a) fired What its at speed an angle at pointof A? Use released from rest at a height h = 3.50R. model to find (a) the maximum each mass ball is and 5.00 g? (b) the total mechaniball Earth sysum height for at the cannon. des without fricp-the-loop h R (Fig. s released from 3.50R. (a) What Figure P8.5 A bead slides without friction around a loop-the-loop. The bead is (b) How large is the normal force on the bead at point A if its

22 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 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 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 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

23 Example 8.7: Box liding Down an Incline Let the ground level represent the zero of gravitational potential. Initially, U i = mgh = (3)g(0.5) = 1.5g and finally U f = 0. K + U + E int = 0

24 Example 8.7: Box liding Down an Incline Let the ground level represent the zero of gravitational potential. Initially, U i = mgh = (3)g(0.5) = 1.5g and finally U f = 0. K + U + E int = 0 K = U f k d = 1.5g (5)(1) = 9.7 J

25 Example 8.7: Box liding Down an Incline Let the ground level represent the zero of gravitational potential. Initially, U i = mgh = (3)g(0.5) = 1.5g and finally U f = 0. K + U + E int = 0 K = U f k d = 1.5g (5)(1) = 9.7 J 2K v = m 2(9.7) = 3 = 2.54 ms 1

26 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 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 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 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

27 Example 8.7 How far does the crate slide on the horizontal floor if it continues to experience a friction force of magnitude 5.00 N? K = E int mv 2 i = f k s s = 1 mvi 2 2f k s = 1 2(5 N) (3 kg)(2.54 ms 1 ) 2 s = 1.94 m

28 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 What If? The block of mass m 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

29 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, # A small block of mass m g is released from rest at point along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R 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

30 Mass in a hemisphere Page 240, #44 R 2R/3 Figure P8.43 Problems 43 and What If? The block of mass m 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

31 Mass in a hemisphere a) K = 1 2 mv 2 = J.

32 Mass in a hemisphere a) K = 1 2 mv 2 = J. b) The system (when including the internal degrees of freedom) is isolated, so: E int = E mech

33 Mass in a hemisphere a) K = 1 2 mv 2 = J. b) The system (when including the internal degrees of freedom) is isolated, so: E int = E mech = U K = (0 mgh) (K f 0) = (0.2)(9.8)(0.30) (0.225) = J

34 Mass in a hemisphere a) K = 1 2 mv 2 = J. b) The system (when including the internal degrees of freedom) is isolated, so: E int = E mech = U K = (0 mgh) (K f 0) = (0.2)(9.8)(0.30) (0.225) = J c) Normally, E int = f k s = µ k ns n is changing, so you can t just multiply f k by s, you have to be savvy. (But you can do it.)

35 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