If a force is applied to an object, the object may experience a change in position, i.e., a displacement.

Transcription

1 If a force is applied to an object, the object may experience a change in position, i.e., a displacement. CHAPTER 6 When a net force is applied over a distance, say, to move an object, mechanical work is done. WORK AND ENERGY Work and kinetic energy work-kinetic energy theorem Work done by a variable force the dot product Power The work done is W = F x Δx, where F x = Fcosθ and Δx = (x x ). x θ " F Δx F x Δx x Potential energy conservative forces non-conservative forces conservative forces and the potentialenergy function equilibrium W = (ma x )Δx. But v = v + ax Δx a x Δx = 1 (v v ). W = 1 mv 1 mv. kinetic energy (K) This is known as the work-kinetic energy theorem: W = K f K i = ΔK.

2 Dimension of work: Units: N m Joule (J). F [M][L] [T] W [M][L] [T] and Δx [L]. (scalar). Question 6.1: A constant force of 10 N acts on a box of mass.0 kg for 3.0 s. If the box was initially at rest and the coefficient of kinetic friction between the box and the surface is µ k = 0.3, (a) what is its speed of the box after 3.0 s? Dimension of kinetic energy: K [M] [L] [T] (scalar). Note: work and kinetic energy have the same dimensions and units. (b) How far did the box travel? (c) What was the work done by the applied force? (d) How much work was done by the frictional force?

3 N F = 10N m =.0 kg f k F Δx v = 0 t = 3.0 s v =? mg (a) The net force acting on the box is F net = F f k = F µ k N = F µ k mg = 10 N (0.3.0 kg 9.81 m/s ) = 4.11 N. Therefore, the acceleration of the box is a = F net m = (4.11 N) (.0 kg) =.06 m/s. The final velocity after 3.0 s is given by v = v + at =.06 m/s 3.0s = 6.18 m/s. (b) To find the distance traveled we have: v = v + aδx, i.e., Δx = v v = ( 6.18 m/s ) a = 9.7 m..06 m/s (c) The work done by the applied force is W = F.Δx = (10 N)(9.7 m) = 9.7 N m. (d) The work done by the frictional force is W f = µ k mg.δx = (0.3)(.0 kg)(9.81 m/s )(9.7 m) = 54.6 J (or N m). Note the net work done by the applied force and the frictional force is W net = ( ) = 38.1 J. It is the net work that produces the increase in the speed of the box as we can see from the work-kinetic energy theorem. W net = ΔK = 1 mv 1 mv = 1 (.0 kg)(6.18 m/s) = 38. J. (Rounding error.)

4 The work-kinetic energy theorem is very important in physics DISCUSSION PROBLEM [6.1]: An object initially has kinetic energy K (= 1 mv ). If it When F and Δx are in the same direction, the work is positive and the velocity increases, because the force F produces a positive acceleration (a = F m ). When F and Δx are in opposite directions the work is negative. For example, when kinetic friction is present, the work done by the frictional force is negative, i.e., f k Δx = ( µ k mg)δx, as it tends to slow the motion. So work can be positive or negative. now moves in the opposite direction with three times its initial velocity, what is its kinetic energy now? A: K B: 3K C: 3K D: 9K E: 9K

5 Let the mass of the motorcycle and rider be M. The work done is given by the work-kinetic energy theorem, i.e., W = ΔK = 1 M v f v i. Question 6.: Which of the following requires the most work to be done by the engine of a motor cycle? A: Accelerating from 60 km/h to 80 km/h. B: Accelerating from 40 km/h to 60 km/h. C: Accelerating from 0 km/h to 40 km/h. D: Accelerating from 0 to 0 km/h. E: The same in each case. For A: ΔK = 1 M ( ) =1400M. For B: ΔK = 1 M ( ) =1000M. For C: ΔK = 1 M ( 40 0 ) = 600M. For D: ΔK = 1 M ( ) 0 = 00M. So case A requires most work by the engine. Even though the change in speed is the same in each case, the work done is greatest for A because the initial speed is the greatest also.

6 F y = 80.0 N 6.0 kg 3.0 m F y mg (a) The work done by the applied force F y is = F y.δy = 80.0 N 3.0 m = 40 J. Question 6.3: A box, of mass 6.0 kg, is raised a distance of 3.0 m from rest by a vertical force of 80.0 N. Find (a) the work done by the force, (b) the work done by the gravitational force, and (c) the final velocity of the box. (b) The work done by the gravitational force is ( mg).δy = (6.0 kg 9.81 m/s ) 3.0 m = 177 J. It is < 0 as the gravitational force is in an opposite direction to the displacement. (c) The net work done on the box is W net = F y.δy mg.δy = 40 J 177 J = 63.0 J. Using the work-ke therorem, W net = ΔK = 1 mv 1 0 mv = 63.0 J, i.e., v = 63.0 J 6.0 kg = 1.0 ( m/s ), v = 4.58 m/s.

7 If F is not parallel to the displacement, we have to use the component of the force parallel to the displacement: θ F i.e., the work done W = ( F cosθ) Δ s = F Δ s cosθ. θ B A Δ s The dot or scalar product of two vectors, A and B is defined as A B = A B cosθ. If A (A x, A y,a z ) and B (B x,b y,b z ) Then A B = A x B x + A y B y + A z B z ~ see revision notes on website ~ Question 6.4: Given two vectors A = (3ˆ i 6ˆ j ) and B = ( 4ˆ i + ˆ j ), calculate (a) the dot product, A B, and (b) the angle between A and B. Therefore, the work done is W = F Δ s

8 (a) Given A = (3ˆ i 6ˆ j ) and B = ( 4ˆ i + ˆ j ), A B = A x B x + A y B y. A B = 3 ( 4) + ( 6) = 4. What does a negative product imply? (b) Also A B = A B cosθ cosθ = A B A B, where θ is the angle between A and B. Now A = 3 + ( 6) = 6.71, B = ( 4) + () = θ = cos 1 A B A B = cos = 143". B ˆ j θ A ˆ i Consider something simple like catching a ball. F BH Note that the ball applies a force on the hand and the hand applies an equal and opposite force on the ball, i.e., F BH = F HB. Assume that in catching the ball the hand moves through a displacement Δ x. Then W BH = F BH Δ x = F BH Δx, and W HB = F HB Δ x = F BH Δ x = W BH. So work can be positive (work done by the hand ON the Δ x F HB ball in slowing it down) and negative (work done by the ball ON the hand). Conventionally, when you do work on a system, we take the work as positive work.

9 OK for a constant force. But what if the applied force is not constant? DISCUSSION PROBLEM [6.]: F F x Δx F x.δx x 1 x x In which of the following is zero net work being done? Work done W = F x Δx = shaded area. A: A ball rolling down a hill. B: A physics student stretching a spring. C: A projectile in free fall towards Earth. D: A box being pulled across a rough floor at constant speed. E: A hockey puck slowing down on the ice. F variable force F xi x Δx 1 i x Work done over the small displacement Δx i is F xi Δx i. W x1 x = Limit Δx i 0 x ( i F xi Δx i ) = F x i.e., area under curve between x 1 and x. x 1 dx,

10 Alternatively, if the force varies in a simple way with position, we can count squares to find the area or use simple geometry. For example... a F x. dx 0 = A 1 + A + A 3. F x (N) A 1 A A 3 x(m) 0 a A 1 = 7.5 N m, A = 5 N m, A 3 = 5 N m. Total area = 37.5 N m Question 6.5: A force, F x = 0.5x 3 N, acts on a object. What is the work done on the object if it is moved from x = 1.5 m to x = 3.0 m? Solve the problem (a) using calculus, and (b) graphically. (a) Plot the function and count squares. F x (N) Approximate number of squares: = But area of a square 0.5 m 1 N = 0.5 N m (b) By definition W = 5 W = 9.4 J. x=3.0 F x x=1.5 F x = 0.5x dx = 0.5 x 3 dx = 0.5 [ ] 4 x4 3.0 = 9.49 J x(m)

11 To determine the final speed we need to determine the work done and then use the work-kinetic energy theorem. The work done is the area under the plot. Question 6.6: A force, F x, shown in the plot, acts on a particle of mass kg, at rest at x = 0. F x (N) x(m) What is the particle s speed at x = 10 m? F x (N) The total area is [ 1] + [] + [3] x(m) = ( ) N m = 46 N m. Therefore, using the work-kinetic energy theorem ΔK = 1 m v f v ( i ) = 46 N m. But v i = 0. v f = ΔK m = 46 N m = 6.78 m/s. kg

12 Interesting cases: (1) the work done lifting a book... h h The net force on the book is: F m g = m a y. The incremental work done is dw = F d y = Fdy. So the total work done is W = Fdy = (mg + ma y )dy = mg dy + m a y dy. 0 h F m g h 0 dy h But, m a y dy = m dv 0 0 dt dy = m dy dv = m vdv dt = 1 mv v v" v v " h 0 = 0, if v " = v (= 0), i.e., the book starts and finishes at rest (so ΔK = 0). h W = mg dy = mgh. 0 Therefore, the work done against the gravitational force, does not depend on how the fast/slow the book is lifted, but only on the height v h 0 v " () work done by the gravitational force on a block sliding down a slope... h As the block slides down the slope, with no friction, the only force on the block in the x-direction is due to the x component of the gravitational force: F x = mg sin θ. The work done by the gravitational force on block is: W = F x.l = mglsin θ. But lsin θ h the height of the slope. W = mgh. Note that W is independent of angle θ; the work done by the gravitational force depends only on the vertical height not the length of the actual path θ l mg θ N mg y x

13 (a) The work done from A B is B W AB = F d s = A F x + B ( F y ) dxˆ A ( i + dyˆ j ) Question 6.7: A force, F = x ˆ i + 3yˆ j ( ) N, where x and y are in meters, acts on a object of mass 5.00 kg. (a) What is the work done on the object if it is moved from point A, ( 1 m, m), to point B, ( m,3 m)? (b) If the speed of the particle at point A is 1.00 m/s, what is its speed when it reaches point B? x B y B = F x.dx + F y.dy. x A y A W AB = x 3.dx + 3y.dy = x3 + 3 y = = 1. J. (b) Using the work energy theorem: W AB = ΔK = K B K A, i.e., K B = W AB + K A. K B = 1. J + 1 ( 5.00 kg ) 1.00 m/s ( ) = 14.7 J 3 = kg ( )v B. v B = 14.7 J 5.00 kg =.4 m/s.

14 POWER DISCUSSION PROBLEM [6.3]: F F A B M M Start Finish Two wind-driven ice-sleds A and B, with masses M and M, respectively, are in a race across a frictionless surface. If the wind exerts a constant force (F) on each boat, which one (if either) wins the race? A: Sled A B: Sled B C: Ha It s a dead-heat Power is the rate at which work is done or energy is dissipated (i.e., quickly or slowly). For example, running up a flight of stairs quickly requires more power than walking up the same flight of stairs slowly. At any instant, the instantaneous power is P = dw F d s = = F d s dt dt dt = F v (scalar), so power can be > 0 or < 0 depending on the angle between F and v. Dimensions: Power Units: J/s watts (W) work done time = [M][L] [T] W 1 kw 1 HP 746 W [M][L] [T] [T]

15 v " v = 0 f k N Question 6.8: (Re-visiting question 5.5) The driver of a 100 kg car moving at 15.0 m/s is forced to slam on the brakes. The car skids to a halt after traveling a distance of 5.5 m. We found that the coefficient of kinetic friction was µ k = 0.45 and that it took 3.40 s for the car to stop. (a) How much work is done by the frictional force? l x " x (a) Two ways to find the work done by the frictional force: (i) W f = f k.l = (µ k mg)l = 0.45(100 kg)(9.81 m/s )(5.5 m) = J. (ii) W f = ΔK = 1 mv 1 mv " = 1 (100 kg)(15.0 m/s) = J. mg (b) What is the average power dissipated by the frictional force? (b) Average power dissipated by the frictional force is P av = W f t = J 3.40 s = W = 39.7 kw. The power is < 0 as W f < 0, which means that energy is removed from the system (as the car is slowing down).

16 Energy and work are one thing... but power is quite another. Look... one gallon of gas has a certain energy content and so can do a fixed amount of work ~ J. But the power produced when the gas burns can have any value... it depends on how fast it burns A gallon of gas will provide enough power to operate a lawnmower for hours (a few HP). However, a gallon of gas may only provide enough power to operate a jet engine for ~ 1 minute (~ 500 HP). One-dimensional example: P = F v = F x v x = ma x v x i.e., a x = P mv x. So, if the engine operates with constant power output, the resulting acceleration is inversely proportional to the velocity, i.e., as v increases, a decreases. Also P = F x v x = ma x v x = mv x dv x dt = d 1 dt mv x = dk dt, i.e., the instantaneous power at some time t is the rate of change of kinetic energy at that same time.

17 Question 6.9: A motor cycle accelerates from zero to 0 mi/h in 1 s. How long would it take to accelerate from zero to 60 mi/h, assuming the power of the engine is constant? We have just seen that the power (P) is related to the change in kinetic energy ( ΔK) and time by: P = ΔK Δt. So, at constant power, Δt ΔK. But ΔK = 1 Mv f as the motorcycle starts from rest. From 0 0 mi/h: ΔK 1 = 1 M 0 = 00M. From 0 60 mi/h: ΔK = 1 M 60 = 1800M. ΔK = 9ΔK 1, i.e., Δt = 9Δt 1. If Δt 1 = 1 s then Δt = 9 s.

18 Conversion factors: 1 hp = 746 W. 10 hp = 7.46 kw. 1 mi/h = 0.45 m/s 10 mi/h = 4.5 m/s, 0 mi/h = 9.0 m/s, 30 mi/h = 13.5 m/s. We found earlier earlier P = ΔK Δt, i.e., Δt = ΔK P = m P v f v ( i ). Question 6.10: A electric motor supplies constant power of 10 hp to the wheels of buggy of mass 500 kg. If all of the power produces motion, how long does it take the buggy to accelerate from (a) 0 to 10 mi/h, (b) 10 mi/h to 0 mi/h, and (c) 0 mi/h to 30 mi/h? 500 kg = W ( ) ( ) v f v i = v f v ( i ). (a) Δt = ( 4.5 m/s) = 0.68 s. (b) Δt = ((9.0 m/s) (4.5 m/s) ) =.04 s. (c) Δt = ((13.5 m/s) (9.0 m/s) ) = 3.39 s.

19 Question 6.11: A 3.00 kg object starts from rest at x = 0.50 m and moves along the x -axis under the influence of a single force F x = x x, where F x is in Newtons and x is in meters. (a) Find the work done by the force as the object moves from x = 0.50 m to x = 3.00 m. (b) Find the power delivered by the force as the object passes the point x = 3.00 m. (a) Since the object is moving along the x-axis, the work done is: ( ) W = x x.dx = 6.00x x = 6( ) x ( ) 1 ( ) = 3.5 J. (b) The instantaneous power at any point is P = F v. Since the motion is confined to the x-axis, P = F x v x, where F x is the force and v x the instantaneous velocity at x = 3.00 m. To get v x we use the work energy theorem to find the kinetic energy at x = 3.00 m. Then: W = ΔK = K( at 3 m) K( at 0.5 m).

20 POTENTIAL ENERGY Since the object is initially at rest K( at 0.5 m) = 0, then W = K( at 3 m) = 1 mv x i.e., v x = W m = 3.5 J 3.00 kg P = F x v x = ( 3.0) ( 3.0) = 3.96 m/s. ( ) N 3.96 m/s F x (at 3 m) = 35.6 W. So far we have seen that when work is done on a isolated object, it leads to a change in kinetic energy. However, there are situations where work does not lead to a change in kinetic energy. We will look at two examples. 1. Springs: consider a spring inside a toy gun. F 1 F x 1 x You compress the spring by applying two equal and opposite forces F 1 and F. The net force on the spring is zero and so there is no change in the kinetic energy of the spring. However, you have done work on the spring W = F 1 Δ x 1 + F Δ x = FΔx, since F 1 = F = F and Δ x 1 = Δ x = Δx.

21 So, what happened to the work you did on the spring? Clearly, the shape of the spring has changed as evidenced by its change in length In fact, when compressed the spring has stored the work you did as potential energy. When released the spring will transfer the stored potential energy to kinetic energy of the ball.. Raising an object in the Earth s gravitational field. In this scenario we must remember that the object and the Earth are a system; so this is a two-particle system. However, you are h h 1 1 mg W you = mgh W grav = mgh not part of the system; you are an external agent that does work on the system. The work done by you on the system in lifting the object from 1 to through a distance h isw you = mgh. But the object starts and finishes at rest, so ΔK = 0. The object is acted on by two equal and opposite forces - ( ) and the force the Earth the force you apply upward mg applies downward ( mg). So, the net work done is W net = W you + W grav = mgh + ( mgh ) = 0, ΔK = 0. But you did work on the system... if it didn t increase the kinetic energy of the object what happened to the work? h h 1 1 mg h v W you = mgh W grav = mgh 1 Note that at the object is capable of doing work; for example, when released it would fall, increasing its speed, and so it could strike a nail and drive it into the floor The amount of work the object can do is W = mgh, which is the same as the amount of work you, an external agent, did raising the object from 1 to.

22 h v 1 When the object is released from rest, ΔK = W 1 = mgh = 1 mv, i.e., v = gh. In fact, at the object/earth system has stored the work you did as potential energy as it has the potential to return the work, when the object is released. When the object is raised, the work done by the gravitational force on the object is W 1 = ( mg)h = mgh. When it is released, the work done by the gravitational force on the object is: W 1 = +mgh, Thus, the net work done by the gravitational force is 1 mgh mgh 1 = 0. CONSERVATIVE FORCES A force is called conservative if the total work it does on an object around any closed path is zero... a closed path is one where the final displacement is zero. Complementary definition: The work done by a conservative force on an object is zero when the particle returns to its initial position, i.e., when the particle moves around any closed path, no matter the route. If the work done in going from point 1 to point is W 1 then (by definition) the work done in going from point to point 1 with a conservative force is W 1 = W 1 along any path. The work done by a conservative force along any path is determined only by the end points not the route. The gravitational force is a conservative force. The elastic force is a conservative force.

23 What about non-conservative forces? a b If you push a book across a table from A to B, you do work against the kinetic frictional force between the book and the table, i.e., W AB = F k.d A B = mgµ k d A B. Consider two paths a and b. Since b > a, then W b > W a. So the work done against the frictional force does depend on the path Therefore, the frictional force is a nonconservative force. Question 6.1: In a certain region of space, the force on an electron is F = cxˆ j, where c is a constant ( > 0). The electron moves in a counterclockwise direction around a square loop in the xy plane. If the corners of the loop are at (0,0), (l,0), (l, l), (0,l), what is the work done on the electron by the force during one complete trip around the loop? Is the force a conservative or non-conservative force? NOTE: The work done by a non-conservative force is non-recoverable (it produces heat, sound, etc.)

24 (0, l) " F = 0 Along the leg from (0,0) to (l, 0) the work done is l W 1 = F " d " l. 0 " The force ( F = cxˆ j) increases linearly with x, but F " d " l, along that leg, so the dot product is zero. Similarly for the leg from (l,l) to (0,l). Thus, no work is done on the electron as it travels along the legs parallel to the x- direction, i.e., W 1 = W 34 = 0. Along the leg from (l,0) to (l,l) the work done is l W 3 = F " d " l l l = cxˆ j dyˆ j = cl dy = cl. 0 y d " l " F d " l " F (0, 0) d " l (l,0) 0 d " l (l, l) " F = clˆ j x 0 (0, l) " F = 0 y d " l " F d " l " F d " l (l, l) " F = clˆ j (0, 0) d " l (l,0) Along the leg from (0,l) to (0,0) the force is zero, since x = 0. So no work is done on the electron along that leg, i.e., W 41 = 0. Thus, the total work done on the electron for the round trip, starting at (0,0) and ending at (0,0), is W = W 1 + W 3 + W 34 + W 41 = cl. Since the start and end points are the same but the total work done by the force is non-zero, F " must be a nonconservative force. Also, since W > 0, the electron has gained energy as it goes around the loop. x ˆ j ˆ i

25 Non-conservative forces and the work-energy theorem Consider an object falling with air-resistance. There are two forces to consider; the gravitational force (conservative) and the drag force (non-conservative). 1 The total work done is W 1 = W grav + W nc, where W grav is the work done by the conservative gravitational force and W nc is the work done by the nonconservative drag force. By the work-energy theorem W 1 = ΔK = K K 1. Also W grav = ΔU = ( U U 1 ). ( ) + W nc, ( ) K 1 + U 1 K K 1 = U U 1 i.e., W nc = K + U ( ), or W nc = E mech E mech1 = ΔE mech. Therefore, the work done by a non-conservative force is equal to the change in mechanical energy. Conservative forces and the potential energy function If the potential energy of a system has a unique value at every point r, we can define a potential energy function, U( r ), which tells us how potential energy varies with position. One property of a conservative force is that the work done by the force can be expressed as the difference between the initial and final values of the potential energy, i.e., dw = F d s = du, where U is the potential energy function. Hence, the work done by a conservative force equals the decrease in the potential energy from point 1 to point. Therefore, the change in potential energy from point 1 to point is ΔU = U U 1 = F d s. Since from above, du = F d s, then, in one-dimension, du = F x.dx, i.e., F x = du dx. 1

26 Given that F x = du, if we know the potential energy dx function, U(x), we can determine the force F x at any point. Conversely, if we know the functional form of the force F x, we can find the potential energy function, since Examples... [1] Gravitational force: ˆ j m g du = F x.dx, i.e., U = F x.dx. Let us find the potential energy function for the gravitational force. We take the y-direction ( ˆ j ) vertical. Then, from above, the incremental change in potential energy as the object falls is du = F d s = ( mgˆ j ) dyˆ j = ( mg)dy. U(y) = mg dy = mgy + U ", where U " is the reference energy when y = 0. Let s plot this function: U(y) 1 ΔU = U U 1 U Height (y) Note: U U(y), i.e., it has a unique value at y. This is the potential energy function for the gravitational force. Normally, we deal only with differences in potential energy, so the choice of U is entirely arbitrary. Note: the force associated with this potential energy function is F z = du = mg, as dz expected. [] Elastic force: F 1 = kx F = kx x If you compress a spring a distance x in the x- direction, the force, F 1, you exert is given by Hooke s Law, i.e., F 1 = kx, where k is the spring constant. The spring does negative work since the force it exerts on you ( F = kx) is opposite to the displacement x.

27 Therefore, the elastic potential energy function U(x) of the spring is given by: du = F d s = ( F ).dx = kx.dx. U(x) = k x.dx = 1 kx + U ". Here are three examples of equilibrium where du dx = 0: U(x) Equilibrium position: x. Stable equilibrium when U(x ) is a minimum, When there is no displacement, i.e., x = 0, the spring has zero elastic potential energy. x i.e., d U dx > 0. U(0) = 0, so U " = 0. At this point, there is no net force acting on the spring (in U(x) x Equilibrium position: x. the x-direction); we say that the spring is in equilibrium. Unstable equilibrium when This is the definition of equilibrium, which we first came U(x ) is a maximum, across in chapter 4 (Newton s 1st Law). x i.e., d U dx < 0. Equilibrium: At equilibrium F = 0, i.e., du dx = 0, so U(x) is an extremum. However, we must exercise care in using that criterion, since du = 0 does not necessarily dx define a minimum in the potential energy function. U(x) x x Equilibrium position: x. Neutral equilibrium since U(x ) is an inflexion point, i.e., d U dx = 0. x

28 (a) The force associated with the potential function is Question 6.13: The potential energy function of an object, of mass.5 kg, confined to move along the x- axis, is U(x) = 3x x 3, where U(x) is in Joules and x is in meters. If the only force acting on the object is the force associated with this potential energy function, (a) plot the force as a function of position. (b) At what positions is the object in equilibrium and what type of equilibrium is associated with each position? (c) Confirm the result by plotting the potential energy function. F(x) = du(x) dx = 6x + 6x. (b) At equilibrium F(x) = 0, i.e., 6x 6x = 6x(x 1) = 0 Therefore, the equilibrium positions are at x = 0 and x = 1 m. To find the type of equilibrium look at the signs of the seond derivative, d U(x) dx, at x = 0 and x = 1 m. Since U(x) = 3x x 3, then d U(x) dx At x = 0: d U(x) dx At x = 1 m: d U(x) dx F(x) x(m) 1 1 = 6 1x. = 6 (> 0) Stable. = 6 (< 0) Unstable.

29 (c) U(J) 5 U(x) = 3x x x(m) 5 The plot of the potential function has a minimum at x = 1 m where the object is in stable equilibrium, and a maximum at x = 1 m where the object is in unstable equilibrium.