PHYSICS 149: Lecture 17

Transcription

1 PHYSICS 149: Lecture 17 Chapter 6: Conservation of Energy 6.7 Elastic Potential Energy 6.8 Power Chapter 7: Linear Momentum 7.1 A Vector Conservation Law 7. Momentum Lecture 17 Purdue University, Physics 149 1

2 ILQ 1 A force of 5 N is applied to the end of a spring, and it stretches 10 cm. How much "farther" will it stretch if an "additional".5 N of force are applied? A).5 cm B) 5 cm C) 10 cm D) 15 cm Lecture 17 Purdue University, Physics 149

3 ILQ A mass is attached to the bottom of a vertical spring. This causes the spring to stretch and the mass to move downward. Does the potential energy of the spring increase or decrease? Does the gravitational potential energy of the mass increase of decrease? A) PE of spring decreases; PE of mass increases B) both decrease C) both increase D) PE of spring increases; PE of mass decreases Lecture 17 Purdue University, Physics 149 3

4 ILQ 3 A simple catapult, consisting of a leather pouch attached to rubber bands tied to two forks of a wooden Y, has a spring constant k and is used to shoot a pebble horizontally. When the catapult is stretched by a distance d, it gives the pebble a speed v. What speed does it give the same pebble when it is stretched to a distance 4d? A) 16 v B) 4 v C) 4 sqrt(4) v D) 64 v E) sqrt(4) v Lecture 17 Purdue University, Physics 149 4

5 Energy and Work Work: Transfer of Energy by Force W F = F s cosθ θ Kinetic Energy (Energy of Motion) K = 1/ mv Work-Energy Theorem: ΣW NC = ΔK +ΔU Gravitational Potential Energy: mm U grav = mgy = -W 1 grav U = G r Lecture 17 Purdue University, Physics 149 5

6 Hooke s Law The deformation change in size or shape of the object is proportional to the magnitude of the force that causes the deformation. Magnitude: k x Direction: Whether the spring is compressed or stretched, the Hooke s force always points toward its relaxation position (x=0). Spring constant k A characteristic of a spring Unit: N/m x=0 F x =k x to left Lecture 17 Purdue University, Physics 149 6

7 Work by Variable Force W = F x Δx Work is area under F vs x plot Force Spring: F = k x Area = ½ k x =W spring Work Distance Work is the area under the F vs x plot Force Work Distance Lecture 17 Purdue University, Physics 149 7

8 Work and Potential Energy of a Spring The force of a spring F = -kx The work 1 1 W kx f + kx i xi = 0 x f = = 1 1 W = kx x The potential energy 1 x = 0 x = x ΔU = W spring = kx f kxi 1 1 U i = kx f Lecture 17 Purdue University, Physics 149 Lecture 15 page 8

9 ILQ You compress a spring by a distance x and store 10 J of energy. How much energy is stored if you compress the spring a distance x? a) 5 J b) 0 J c) 30 J d) 40 J Lecture 17 Purdue University, Physics 149 9

10 Work Done by an Ideal Spring The work done by a spring as its movable end moves from equilibrium (x i =0) to the final position x f is x f Wspring = FHooke ds xi = 0 beyond the scope x = 0 ( kx) dx 1 = kx f The work done by the spring is the (negative) area under the F x (x) graph. The work done by an ideal spring depends only on the initial and final positions of the moveable end (independent of the path taken). Hooke s force is a conservative force. Potential energy may be defined. = x i f of this class independent of path! Area = ½ base height Lecture 17 Purdue University, Physics

11 Elastic Potential Energy The work done by an ideal spring is independent of path (see previous slide) It means that the spring force is a conservative force Thus, we may define potential ti energy for the spring force. ΔU = U ( x f ) U ( x beyond the scope of this class i ) W = = = 1 x x x i x i Hooke f f F Hooke ds ( kx) dx 1 kx f kx i U x = U x 1 + kx 1 kx ( f ) ( i ) f i If we let U x ) 0 at x = 0, 1 U = ( x f ) kx f ( i i arbitrary choice Lecture 17 Purdue University, Physics

12 ILQ All springs and masses are identical. (Gravity acts down). Which of the systems below has the most potential energy stored in its spring(s), relative to the relaxed position? A) 1 B) C) same (1) () Lecture 17 Purdue University, Physics 149 1

13 ILQ: Solution The displacement of (1) from equilibrium will be half of that of () (each spring exerts half of the force needed to balance mg) 0 d d (1) () Lecture 17 Purdue University, Physics

14 ILQ: Solution The potential energy stored in (1) is 1 kd = kd The potential energy stored in () is ( ) 1 kd = kd The spring P.E. is twice as big in ()! d 0 d (1) () Lecture 17 Purdue University, Physics

15 Vertical Springs A spring is hung vertically. Its relaxed position is at y = 0 (a). When a mass m is hung from its end, the new equilibrium position is y e (b). Recall that the force of a spring is F s = -kx. In case (b) F s = mg and x = y e : (a) k (b) j y = 0 -kyy e - mg = 0 (y e < 0) ) m y = y e mg = -ky e mg -ky e (ok since y e is a negative number) Lecture 17 Purdue University, Physics

16 Vertical Springs The potential ti energy of the spring-mass system is: 1 U = ky + mgy + C (a) k (b) j U = 1 ky but mg = -ky e y = 0 ky e y + C choose C to make U=0 at y=y: y e : = kye kye + C C = ky e y = y e m mg -ky e Lecture 17 Purdue University, Physics

17 So: Vertical Springs 1 1 (a) (b) U= ky kyey+ kye j k 1 ( = ky + y ) e yy e which can be written: U 1 = k( y y ) e mg m -ky e y = 0 y = y e Lecture 17 Purdue University, Physics

18 Vertical Springs U 1 = k ( y ) y e So if we define a new y k coordinate system such that y = 0 is at the equilibrium position, ( y = y - y e ) then we get the simple result: y = 0 m U = 1 ky (a) (b) j Lecture 17 Purdue University, Physics

19 Vertical Springs If we choose y = 0 to be at the equilibrium position of the mass (a) (b) hanging on the spring, we can define the potential in the simple form. k j U = 1 ky Notice that g does not appear in this expression!! By choosing our coordinates and constants t cleverly, we can hide the effects of gravity. m y = 0 Lecture 17 Purdue University, Physics

20 ILQ: Energy Conservation In (1) a mass is hanging from a spring. In () an identical mass is held at the height of the end of the same spring in its relaxed position. Which correctly describes the relation of the potential energies of the two cases? (a) U 1 > U (b) U 1 < U (c) U 1 = U case 1 case d Lecture 17 Purdue University, Physics 149 0

21 ILQ: Solution In case 1, it is simplest to choose the mass to have zero total potential energy (sum of spring and gravitational potential energies) at its equilibrium position. 1 In case the total potential energy is then U = kd kd relaxed y = 0, U 1 = 0 d y = d The answer is (b) U 1 < U. Lecture 17 Purdue University, Physics 149 1

22 Power (Rate of Work) P = W / Δt Units: Joules/Second = Watt W = F. Δr = F Δr cosθ = F (v Δt) cosθ P ΔW = Δt F Δr P=Fvcosθ How much power does it take for a (70 kg) student to run up the stairs (5 meters) in 7 seconds? P = W / t = m g h / t = (70 kg) (9.8 m/s ) (5 m) / 7 s = 490 J/s or 490 Watts Lecture 17 Purdue University, Physics 149 v

23 Example Lars, of mass 8.4 kg, has been working out and can do work for about.0 min at the rate of 746 W. How long will it take him to climb three flights of stairs, a vertical height of 1.0m? As Lars climbs the stairs, he increases his gravitational potential energy. The rate of potential energy increase must be equal to the rate he does work. P av = Δ t Δ E Δ U mg Δ y = = Δt Δt Δt mgδy (8.4kg)(9.80m / = = P 746W av s )(1.0m) = 13.0 s Lecture 17 Purdue University, Physics 149 3

24 Power A 000 kg trolley is pulled up a 30 degree hill at 0 mi/hr by a winch at the top of the hill. How much power is the winch providing? y v θ mg T x winch The power is P = F. v = T. v Since the trolley is not accelerating, the net force on it must be zero. In the x direction: T - mg sin θ = 0 T = mg sin θ Lecture 17 Purdue University, Physics 149 4

25 Power T. P= v = Tv since T is parallel to v v y T x winch So P = mgv sin θ v = 0 mi/hr = 8.94 m/s g = 9.81 m/s m = 000 kg sin θ = sin(30 o ) = 0.5 θ mg and P = (000 kg)(9.81 m/s )(8.94 m/s)(0.5) = 87,700 W Lecture 17 Purdue University, Physics 149 5

26 Power Power is the rate at which energy is transferred, or equivalently, the rate at which work is done (that is, work per a time interval). Average Power: the average rate of energy conversion Instantaneous Power: the instantaneous rate at which a force F does work when the object it acts on moves with velocity v W F Δ r cosθθ P = = = Fv cosθ Δt Δt Power is a scalar quantity. Units: W, J/s, etc. Unit conversion: 1 W = 1 J/s Note that kwh (kilowatt-hour) is a unit of energy, not power. Power is denoted by P. Lecture 17 Purdue University, Physics 149 6

27 ILQ What power must an engine have if it is to be used to raise a 5 kg load 10 m in 4 seconds? a) 5 W b) 65 W c) 1000 W d) 500 W Lecture 17 Purdue University, Physics 149 7

28 Example: The Dart Gun In this case, there are three forces acting on the dart. But, the directions of gravity and normal forces are perpendicular to the displacement of the dart, so the work done by the two forces are zero. Hooke force (a conservative force) is the only force which does work, so the mechanical energy is conserved. E mech = K i + U i = K f + U f = const (b/c W nc =0) K i = 0 (b/c v i = 0) U = i ½kx i K f = ½mv f U f = 0 (b/c x f = 0) 0 + ½kx i = ½mv f + 0 Thus, v f = sqrt(k/m) x i = 11 m/s Lecture 17 Purdue University, Physics 149 8

29 Work-Energy Key Ideas Σ F = m a multiply both sides by d Σ F d = m a d (note: a d = ½ Δv ) Σ F d = ½ m Δv Σ W = ΔK K Define Work and Kinetic Energy Impulse-Momentum Σ F = m a multiply both sides by Δt Σ F Δt = m a Δt (note: a Δt = Δv) Σ F Δt = m Δv Σ I = Δp Define Impulse and Momentum Lecture 17 Purdue University, Physics 149 9

30 Momentum is Conserved Momentum is Conserved meaning it can not be created nor destroyed Can be transferred Total Momentum does not change with time Momentum is a VECTOR 3 Conservation Laws in one! Lecture 17 Purdue University, Physics

31 A Vector Conservation Law When a vector quantity is conserved in an interaction, both its magnitude and direction are unchanged (or equivalently, all components are unchanged). Lecture 17 Purdue University, Physics

32 Example: Momentum What is the momentum of an automobile (weight = 9800 N) when it is moving at 35 m/s to the south? W = mg m = W/g = (9800 N) / (9.8 m/s ) = 1000 kg p = mv = (1000 kg) (35 m/s south) = 35,000 kg m/s south Don t forget that momentum p is a vector! (We need both its magnitude and direction.) Lecture 17 Purdue University, Physics 149 3

33 Pushing Off Fred and Jane are on skates facing each other. Jane then pushes Fred with force F NL Fred: F JF = m Fred a Δv Fred = a Δt = (F JF /m Fred ) Δt m Fred Δv Fred = F JF Δt NL Jane: F FJ = m Jane a Δv Jane = a Δt = (F/m Jane ) Δt a = Δv/Δt m Jane Δv Jane = F JF Δt Jane Jane JF N3L: For every action, there is an equal and opposite reaction. F FJ =-F JF m Fred Δv Fred = -m Jane Δv Jane Lecture 17 Purdue University, Physics

34 Pushing Off Fred (75 kg) and Jane (50 kg) are on skates facing each other. Jane then pushes Fred with a constant force F = 45 N for a time Δt = 3 seconds. Who will be moving fastest at the end of the push? A) Fred B) Same C) Jane Fred F = +45 N (positive direct.) I = Ns = 135 Ns Jane F = -45 N Newton s 3 rd law I = Ns = -135 Ns I = Δp I = Δp = mv f mv i = mv f mv i I/m = v f -v i v f = 135 N-s / 75 kg I/m = v f -v i v f = -135 N-s / 50 kg = m/s = m/s Note: P fred + P jane = (1.8) 75 + (-.7) 50 = 0! Lecture 17 Purdue University, Physics