Physics 8 Wednesday, October 2, 2013

Transcription

1 Physics 8 Wednesday, October 2, 2013 Today: Finish forces. Start work. HW5 due Friday. Read rest of Ch10 for Monday. I re-posted mazur ch10.pdf with figures repaired sorry. Short (6 Q s) HW6 due next Wednesday (covers Ch9). After ch9, the pace of reading will slow down to 1ch/wk! 10-min quiz today on a modified HW3 problem. 1pg notes ok HW help sessions: Wed 7pm DRL 2N36, Thu 7pm DRL 3W2. (I ll figure out a special case for next week.) I think every equation needed for doing HW appears here: positron.hep.upenn.edu/wja/phys008_2013/equations.pdf Also note key equations, worked example problems, etc., in study guide chapters on the Canvas reading page. Good luck with your ARCH midterm reviews!

2 Sorry: Monday s blackboard algebra was totally unclear (The answer made so much sense to me by intuition alone that I underestimated the steps involved in solving it systematically!) When a 5.0 kg box is suspended from a spring, the spring stretches to 1.0 m beyond its equilibrium length. In an elevator accelerating upward at 0.98 m/s 2, how far will the spring stretch with the same box attached? (A) 0.50 m (B) 0.90 m (C) 1.0 m (D) 1.1 m (E) 1.2 m (F) 1.9 m (G) 2.0 m

3 Since springs are newer for us than gravity, let s choose the x axis to point downward, so that x x 0 for the spring increases as the spring stretches. The force exerted by the spring on the box is F x,sb = k(x x 0 ) The gravitational force exerted by Earth on the box is Fx,Eb G = +mg The equation of motion for the box is given by Newton s 2nd law: ma x,b = F x,sb + Fx,Eb G = mg k(x x 0) Solving for x x0 (i.e., how far we stretch the spring) gives k(x x 0 ) = m(g a x ) (x x 0 ) = m k (g a x) (free-body diagram on next page)

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5 With x axis pointing downward, we got (x x 0 ) = m k (g a x) In situation #1, a x = 0 and x x 0 = 1.0 m. (x 1 x 0 ) = m k (g a x) 1.0 m = m k (9.8 m/s2 ) m k = s2 In situation #2, a x = 0.98 m/s 2 (because we defined upward to be negative) and we need to find x x 0. (x 2 x 0 ) = m k (g a x) = s2 ( 9.8 m s m s 2 ) = 1.1 m

6 Quicker way: use ratio of two equations With x axis pointing downward (so upward is negative), we got (x 1 x 0 ) = m k (g a x) In situation #1, a x = 0 and x x 0 = 1.0 m. In situation #2, a x = 0.98 m/s 2 ( upward is negative) and we want x 2 x 0. Taking the ratio of the situation #2 and #1 equations, So then x 2 x 0 = (g a x2) x 1 x 0 g 0 x 2 x m = x 2 x 0 = (1.1)(1.0 m) = 1.1 m = 1.1

7 Even quicker way The equation of motion for the box is ma x = (vector sum of forces) x If we let the x axis point upward (which seems more natural), then we can write the equation of motion as ma x = mg + F x,by spring Then rearrange to separate out what the spring does: m(a x + g) = F x,by spring So now the ratio of spring forces in the two situations is m(a x + g) m(0 + g) = k(x 2 x 0 ) k(x 1 x 0 ) = 1.1g g

8 Tension: Suppose a horse can pull 1000 N F A on rope = F rope on A F B on rope = F rope on B F A on rope = F B on rope = 1000 N F rope on A = F rope on B = T T = 1000 N Because the horse s feet don t slip (ground/friction pushes each horse away from the other horse, opposing force of rope on horse) a = 0

9 Ideal tree, wall, etc. stays put, no matter how hard I pull F A on rope = F rope on A F rope on tree = F tree on rope F A on rope = F rope on A = T = F rope on tree = 1000 N T = 1000 N Because the ground applies to an ideal tree/wall/etc. whatever force is needed to keep the ideal tree/wall/etc. from moving, a tree = 0

10 Tree stays put, no matter how hard I pull F A+B on rope = F rope on A+B 2000 N = F A+B on rope = F rope on A+B = T F rope on tree = F tree on rope And again, the ground pulls on the ideal tree in such a way that a tree = 0

11 Horse C loses his footing when he pulls > 1000 N 2000 N = F A+B on rope = F rope on A+B = T = F rope on C T = 2000 N Force of ground on C is 1000 N to the right. Tension in rope pulls on C 2000 N to the left. C accelerates to the left. a C = (2000 N 1000 N)/m (What happens next depends on whether rope stays taut...)

12 In the 17th century, Otto von Güricke, a physicist in Magdeburg, fitted two hollow bronze hemispheres together and removed the air from the resulting sphere with a pump. Two eight-horse teams could not pull the halves apart even though the hemispheres fell apart when air was readmitted. Suppose von Güricke had tied both teams of horses to one side and bolted the other side to a heavy tree trunk. In this case, the tension on the hemispheres would be (A) twice (B) exactly the same as (C) half what it was before.

13 Chapter 9 Work 1. If you graph the work done by a force on an object as a function of the object s position, what graphical feature represents the force exerted on the object? The slope represents the force exerted on the object. 2. When you stand up from a seated position, you push down with your legs. So then do you do negative work when you stand up? Because the floor does not experience any displacement, you aren t performing any work on the floor. Therefore you re not doing negative work. No in this case, the force exerted by me is downward, but I m also pushing the earth downward and away from me. Since the direction of the force and force displacement are the same, the direction of work is positive.

14 Chapter 9 Work I guess this question depends on what one considers to be the system. When standing up, legs bolster us up in the same direction as the contact force, against the force of gravity. There is no negative work involved. (My legs do positive work on my body.)

15 Reading question 2 had no really simple answer When you stand up from a seated position, you push down with your legs. Does this mean you do negative work when you stand up? Suppose system = me + earth + floor + chair K = 0 U = mg ( x) my c.o.m. > 0 E thermal = 0 (debatable but irrelevant) E source = mg ( x) my c.o.m. < 0 W = 0 There are no external forces. Everything of interest is inside the system boundary.

16 Let s try choosing a different system. When you stand up from a seated position, you push down with your legs. Does this mean you do negative work when you stand up? Suppose system = me + floor + chair K = 0 U = 0 (U G undefined if Earth not in system) E thermal = 0 (debatable but irrelevant) E source = mg ( x) my c.o.m. < 0 W = mg ( x) my c.o.m. < 0 External force of gravity does negative work on me. Point of application of this external force is my body s center of mass.

17 Let s try answering a slightly different question. When a friend stands me up from a chair (e.g. my knees are weak today), does my friend do positive or negative work? Suppose system = me + earth + floor + chair K = 0 U = mg ( x) my c.o.m. > 0 E thermal = 0 (debatable but irrelevant) E source = 0 W = mg ( x) my c.o.m. > 0 My friend applies an upward force beneath my arms. The point of application of force is displaced upward.

18 Let s include my friend as part of the system. When a friend stands me up from a chair (e.g. my knees are weak today), does my friend do positive or negative work? Suppose system = me + earth + floor + chair + friend K = 0 U = mg ( x) my c.o.m. > 0 E thermal = 0 (debatable but irrelevant) E source = mg ( x) my c.o.m. < 0 W = 0 There is no external force. Everything is within the system.

19 Back to the original reading question When you stand up from a seated position, you push down with your legs. Does this mean you do negative work when you stand up? I think the work done ON the system BY external forces is either positive (if my legs are considered external to the me+earth+floor system and are supplying the energy to lift me) or zero (if my legs are part of the system).

20 Some equation sheet entries for Chapters 8+9 positron.hep.upenn.edu/wja/phys008_2013/equations.pdf Work (external, nondissipative, 1D): W = F x (x) dx which for a constant force is G.P.E. near earth s surface: U gravity = mgh Force of gravity near earth s surface (force is du gravity dx ): W = F x x Power is rate of change of energy: P = de dt Constant external force, 1D: P = F x v x F x = mg Potential energy of a spring: U spring = 1 2 k(x x 0) 2 Hooke s Law (force is du spring dx ): F by spring ON load = k(x x 0 )

21 A few key ideas from Chapters 8 and 9 Impulse (i.e. momentum change) delivered by external force: force = d(momentum) dt J = F external dt External force exerted ON system: force = d(work) W = dx F x dx Force exerted BY spring, gravity, etc.: d(potential energy) force = dx E system = flow of energy into system = work done ON system: work = (energy) = K + U + E source + E thermal Notice that work : energy :: impulse : momentum

22 Suppose you want to ride your mountain bike up a steep hill. Two paths lead from the base to the top, one twice as long as the other. Compared to the average force you would exert if you took the short path, the average force you exert along the longer path is (A) one-fourth as large. (B) one-third as large. (C) one-half as large. (D) the same. (E) twice as large. (F) undetermined it depends on the time taken

23 A piano mover raises a 100 kg piano at a constant rate using the pulley system shown here. With how much force is she pulling on the rope? (Ignore friction and assume g 10 m/s 2.) (A) 2000 N (B) 1500 N (C) 1000 N (D) 750 N (E) 500 N (F) 200 N (G) 50 N (H) impossible to determine.

24 Block and tackle: mechanical advantage

25 Believe it or not, this relates to architecture! My friend and I both want to hang on to a rope by our hands, perhaps because being up above the ground lets us peek over a tall fence and see into an amazing new construction site next door. We consider two different methods of hanging onto the rope. In the first method, I hold the rope with my hands, about 5 meters off the ground, and my friend (whose mass is the same as mine) holds the rope with his hands, about 3 meters off the ground. In the second method, I told the rope with my hands, as before, and my friend holds onto my feet (instead of the rope). Let s draw a picture, to make it more clear.

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27 The downward force of my hands on the rope is... (A) The same for both methods: equal to mg (m = my mass) (B) The same for both methods: equal to 2mg (C) Twice as much for 1st method (2mg vs. mg) (D) Twice as much for 2nd method (2mg vs. mg)

28 Kansas City Hyatt Regency skywalk collapse For more like this, read To Engineer is Human by Henry Petroski.

29 As designed, each of the two skywalks hangs onto the rope with its own hands. As built, the lower skywalk s hands are effectively hanging onto the upper skywalk s feet! So the upper skywalk s grip on the rope feels 2 larger force than in original design. Oops!

30 Look! A real use for free-body diagrams! The author uses the symbol P for a point force (or point load), as is the custom in engineering. When you see P here, pretend it says F or mg instead.

31 Upper skywalk loses its grip on the rope

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33 A spring-loaded toy dart gun is used to shoot a dart straight up in the air, and the dart reaches a maximum height of 8 m. The same dart is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the dart go this time (neglecting friction)? (A) 1 m (B) 2 m (C) 4 m (D) 8 m (E) 16 m (F) 32 m

34 Stretching a certain spring 0.10 m from its equilibrium length requires 10 J of work. How much more work does it take to stretch this spring an additional 0.10 m from its equilibrium length? (A) No additional work (B) An additional 10 J (C) An additional 20 J (D) An additional 30 J (E) An additional 40 J

35 At the bowling alley, the ball-feeder mechanism must exert a force to push the bowling balls up a 1.0 m long ramp. The ramp leads the balls to a chute 0.5 m above the base of the ramp. About how much force must be exerted on a 5.0 kg bowling ball? (A) 200 N (B) 100 N (C) 50 N (D) 25 N (E) 5.0 N (F) impossible to determine.

36 A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom. To achieve a speed 2v at the bottom, how many times as high must a new ramp be? (A) 1 (B) (C) 2 (D) 3 (E) 4 (F) 5 (G) 6

37 Spring B is stiffer than spring A. Which one has more energy stored in it if you (1) compress both springs with the same force and (2) compress both springs the same displacement from their equilibrium lengths? (A) Same energy is stored in both cases (B) Stiffer spring stores more energy in both cases (C) Stiffer spring stores less energy in both cases (D) Stiffer spring stores more energy in case (1), less energy in case (2) (E) Stiffer spring stores less energy in case (1), more energy in case (2)

38 Suppose you drop a 1 kg rock from a height of 5 m above the ground. When it hits, how much force does the rock exert on the ground? (Take g 10 m/s 2.) (A) 0.2 N (B) 5 N (C) 50 N (D) 100 N (E) impossible to determine.

39 HW5 due Friday. Read 2nd half (equations) of Ch10 for Mon. Updated CH10 PDF on Canvas fixed missing figures. HW help sessions: Wed 7pm DRL 2N36, Thu 7pm DRL 3W2. positron.hep.upenn.edu/wja/phys008_2013/equations.pdf Good luck with your ARCH midterm reviews! Quiz on HW #3 (handing out now) 10 minutes. Work alone. Closed book, 1 page of handwritten notes OK. Calculator OK. Relax: 5% weight, 80% 100%. A load of coal, of inertia kg is dropped vertically from a bunker into a railroad hopper car of inertia kg. The car is initially coasting to the right at 2.0 m/s on a level track, with negligible friction. (a) What is the velocity of the car after the coal falls into the car? Show your work. State magnitude and direction (e.g. right or left). (b) Would you describe this collision as elastic, inelastic, totally inelastic, or explosive? State your reasoning.