Lecture 5 - notes PDF

Preview

Summary

notes...

Description

Chapter 6 Work and Energy • Work Done by a Constant Force • Work Done by a Varying Force • Kinetic Energy, and the Work-Energy Principle • Potential Energy • Conservative and Nonconservative Forces • Mechanical Energy and Its Conservation • Problem Solving Using Conservation of Mechanical Energy • Energy Conservation with Dissipative Forces: Solving Problems • Power

1

6-1 Work Done by a Constant Force In the past few chapters we used Newton’s Laws of motion to describe dynamics. In principle we can use these laws to solve any problem in mechanics. However, in many situations due to the vector nature of the equations it maybe hard to implement these laws to solve problem. Here we introduce an equivalent approach to mechanics. We introduce the ideas of work and energy. The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of displacement: F

F

θ

θ

d

W = F d cos θ In the SI system, the units of work are joules: J = N · m. Solving work problems: 1. Draw a free-body diagram. 2. Choose a coordinate system. 3. Apply Newton’s laws to determine any unknown forces. 4. Find the work done by a specific force. 5. To find the net work, either (a) find the net force and then find the work it does, or (b) find the work done by each force and add. 2

(1)

Examples: 1) A person carries a box of mass m and moved it horizontally a distance d. How much work did she do? FS

d

mg

Wg = mgd cos(90) = 0 WFS = Fs d cos(90) = 0 WP = FP d cos(0) = 0. (no force is applied horizontally, FP = 0) 2) A mass is moving in a uniform circular motion. How much work is done by the centripetal force as the mass move a distance d along the circle? v m

v

d

FR

m FR

WFR = FR d cos(90) = 0 Since FR is always perpendicular to the displacement. 3

3) A box of mass m = 10 kg is pulled by a force of 100 N for a distance of d = 10 m across a floor of kinetic friction µk = 0.25. The force is acting at an angle of 37o above the horizontal. Calculate the work done by each force. Free−body diagram F = 100 N FN 37o Ffr

mg

WF = F d cos(θ) = 100 × 10 × cos(37) = 798.6J Wg = mgd cos(90) = 0 WN = FN d cos(90) = 0 X Fiy = F sin θ + FN − mg = 0 → FN = mg − F sin θ i

Ff r = µk FN = 0.25(98 − 100 sin(37) = 9.45N

WFf r = Ff r d cos(180) = −94.5J

A box of mass 4.0 kg is accelerated from rest by a force across a floor at a rate of 2.0 m/s 2 for 7.0 s. Find the net work done on the box. Solution Assume that the box starts from rest. 1 1 d = x − x0 = v0 t + at2 = 0 + 2.0 × 7.02 = 49 m. 2 2 Then the work done in moving the crate is W = F d cos 0◦ = mad = (4.0 kg)(2.0 m/s2 )(49 m) = 390 J 4

(6-10) A 380-kg piano slides 2.9 m down a 25◦ incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-36). Determine: (a) the force exerted by the man, (b) the work done on the piano by the man, (c) the work done on the piano by the force of gravity, and (d) the net work done on the piano. Ignore friction. y FN

x

FP mg

θ

Solution (a) Write Newtons second law on each direction for the piano, with an acceleration of 0. X Fy = FN − mg cos θ = 0 ⇒ FN = mg cos θ X Fx = mg sin θ − FP = 0 ⇒ FP = mg sin θ = (380)(9.80)(sin 25◦ ) = 1574 N (b) The work done by the man is the work done by FP . The angle between FP and the direction of motion is 180◦ WP = FP d cos 180◦ = −(1574)(2.9) = −4565 J. (c) The angle between the force of gravity and the direction of motion is 65◦ . Calculate the work done by gravity. WG = FG d cos 63◦ = mgd cos 63◦ = (380)(9.80)(2.9) cos 65◦ = 4565 J. (d) Since the piano is not accelerating, the net force on the piano is 0, so the net work done on the piano is also 0. This can also be seen by adding the two work amounts calculated. Wnet = WP + WG = −4565 J + 4565 J = 0 5

(6-12) A grocery cart with mass of 16 kg is being pushed at constant speed up a 12◦ ramp by a forceFP which acts at an angle of 17◦ below the horizontal. Find the work done by each of the forces (mg, FN , FP ) on the cart if the ramp is 7.5 m long Solution Consider a free-body diagram for the grocery cart being pushed up the ramp. If the cart is not accelerating, then the net force is 0 in all directions. The work done by the normal force is 0 since the normal force is perpendicular to the displacement. The angle between the force of gravity

FN

y x

FP φ θ θ

mg and the displacement is 90◦ + θ = 102◦ . The angle between the normal force and the displacement is 90◦ . The angle between the pushing force and the displacement is φ + θ = 29◦ mg sin θ cos(φ + θ ) W mg = mgdcos112◦ = (16)(9.80)(7.5) cos 102◦ = −244.5 J X

Fx = FP cos(φ + θ) − mg sin θ − FP = 0 ⇒ FP =

Wnormal = FN d cos 90◦ = 0. mg sin 12◦ d cos 29◦ = mgd sin 12◦ = (16)(9.80)(7.5)sin12◦ = 244.5 J. WP = FP d cos 29◦ = cos 29◦

6-2 Work Done by a Varying Force For a force that varies, the work can be approximated by dividing the distance up into small pieces, finding the work done during each, and adding them up. As the pieces become very narrow, the work done is the area under the force vs. distance curve. 6

F force

d di

distance

df

∆xi

W

≈

W →

X

F (xi )∆xi

(2)

i

lim

∆xi →0

X

F (xi )∆xi

(3)

i

6-3 Kinetic Energy, and the Work-Energy Principle Energy was traditionally defined as the ability to do work. We now know that not all forces are able to do work; however, we are dealing in these chapters with mechanical energy, which does follow this definition. Consider a mass where a constant net force Fnet is applied on it. We know from Newton’s 2nd Law that the object will have a constant acceleration a. Lets calculate the work done by this force in displacing the mass a distance

7

d which we define as Wnet. Wnet = Fnet d Fnet = ma, and a =

v22 − v 21 since a is constant 2d

v22 − v 21 d 2d 1 1 = mv22 − mv12 2 2

Wnet = m Wnet

We define the Kinetic Energy of a mass m moving with a velocity v , 1 KE = mv2 2 This means that the work done is equal to the change in the kinetic energy: Wnet = ∆KE This result is known as Work-Energy Theorem. Even though we derived it for a constant net force it nevertheless is true even if we have a variable force. If the net work is positive, the kinetic energy increases. If the net work is negative, the kinetic energy decreases. Example: A car of mass 2000 kg moving at 40 m/s slows down to 20 m/s. What is the net work done?  1  Wnet = ∆KE = m v22 − v 21 2  2  1 = 2000 20 − 402 2 = −1.2 × 106 J. Lost KE. (6-19) Two bullets are fired at the same time with the same kinetic energy. If one bullet has twice the mass of the other, which has the greater speed and by what factor? Which can do the most work? 8

Solution The kinetic energies of both bullets are the same. Bullet 1 is the heavier bullet. 1 1 KE1 = KE2 ⇒ m1 v12 = m2 v22 2 2 √ ⇒ m1 v12 = m2 v22 ⇒ 2m2 v 12 = m2 v22 ⇒ 2v12 = v22 ⇒ v2 = 2v1 . √ The lighter bullet has the higher speed, by a factor of 2. Both bullets can do the same amount of work. m1 = 2m2

(6-22) If the speed of a car is increased by 50%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver’s reaction time. Solution The work needed to stop the car is equal to the change in the car’s kinetic energy. That work comes from the force of friction on the car. Assume the maximum possible frictional force, which results in the minimum braking distance. Thus Ff r = µs FN . The normal force is equal to the car’s weight if it is on a level surface, so Ffr = µs mg . 1 1 mv22 − mv12 2 2 1 2 v12 ⇒ −µs mgd = 0 − mv1 ⇒ d = 2 2gµk

Wnet = ∆KE ⇒ Ff r d cos 180◦ =

Since d ∝ v12 , if v1 increases by 50%, or is multiplied by 1.5, then d will be multiplied by a factor of (1.5)2 , or 2.25. (6-24) One car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 8.0 m/s, they then have the same kinetic energy. What were the original speeds of the two cars? Solution The first car mentioned will be called car 1. So we have these statements:   1 1 1 1 2 2 KE1 = KE2 ⇒ m1 v1 = m 2 v2 2 2 2 2 KE1, f ast = KE2, f ast

9

Now use the mass information, that m1 = 2m2 .   1 1 1 1 2 2 1.) (2m2 )v 1 = m2 v2 ⇒ v 21 = v 22 ⇒ 2v1 = v2 2 2 2 4 2.) KE1, f ast = KE2, f ast 1 1 2.) (2m2 )(v1 + 8.0)2 = m2 (v2 + 8.0)2 ⇒ 2(v1 + 0.8)2 = (v2 + 0.8)2 2 √ 2 1.) → 2.) ⇒ 2(v1 + 0.8) = (2v1 + 0.8) 8.0 ⇒ v1 = √ m/s = 5.457 m/s, v2 = 2v1 = 11.314 m/s. 2 (6-25) A 265-kg load is lifted 18.0 m vertically with an acceleration a = 0.160g by a single cable. FT

mg

Determine (a) the tension in the cable From the free-body diagram for the load being lifted, write Newton’s second law for the vertical direction, with up being positive. X Fy = FT − mg = ma = 0.16mg ⇒ FT = 1.16mg = 1.16(265)(9.80) N = 3010 N.

(b) the net work done on the load The net work done on the load is found from the net force. W net = Fnet d cos(0) = (0.160mg )d = 0.160(265)(9.80)(18.0)J = 7480 J. 10

(c) the work done by the cable on the load The work done by the cable on the load is as follows: Wcable = FT d cos(0) = (1.160mg)d = 1.160(265)(9.80)(18.0)J = 5.42×104 J (d) the work done by gravity on the load The work done by gravity on the load is as follows: WG = mgd cos(180) = −mgd = −(265)(9.80)(18.0) J = −4.67 × 104 J (e) the final speed of the load assuming it started from rest. Use the work-energy theorem to find the final speed, with an initial speed of 0. 1 1 Wnet = KE2 − KE1 = mv22 − mv12 2 2 r 2Wnet ⇒ v2 = + v 12 = 7.51 m/s. m

6-4 Potential Energy An object can have potential energy by virtue of its surroundings. Potential energy is energy of an object or a system due to position or configuration. Familiar examples of potential energy: • A wound-up spring • A stretched elastic band • An object at some height above the ground

11

y y2 d

Fext

Fext = mg no acceleration

h

mg y1

Gravitational Potential Energy Suppose we raise an object of mass m a height h without accelerating it as shown below. In raising a mass m to a height h, the work done by the external force is, Wext = Fext d cos(0) = mgh = mg(y2 − y1 ) We therefore define the gravitational potential energy: P Eg = mgy This potential energy can become kinetic energy if the object is dropped. Potential energy is a property of a system as a whole, not just of the object (because it depends on external forces). If P Eg = mgy, where do we measure y from? It turns out not to matter, as long as we are consistent about where we choose y = 0. Only changes in potential energy can be measured. We usually choose the lowest potential energy of the system as y = 0. (6-29) A 66.5-kg hiker starts at an elevation of 1270 m and climbs to the top of a peak 2660 m high. Solution 12

(a) What is the hiker’s change in potential energy? The change in gravitational potential energy is given by the following: ∆PEG = mg(y2 − y1 ) = (66.5)(9.80)(2660 − 1270) J = 9.06 × 105 J. (b) What is the minimum work required of the hiker? The minimum work required by the hiker would equal the change in potential energy, which is 9.06 × 105 J. (c) Can the actual work done be greater than this? Explain. Yes. The actual work may be more than this, because the hiker almost certainly had to overcome some dissipative forces such as air friction. Also, while stepping up and down, the hiker does not get the full amount of work back from each up-down event. For example, there will be friction in the joints and muscles.

Elastic Potential Energy Potential energy can also be stored in a spring when it is compressed; the figure below shows potential stored due to stretching or compression. unstreched spring

x=0 compress

strech FS

FA

FA

F

S

x

x

and the force the spring exerts is: Fs = −kx Hooke′ s Law. Here k is called the spring constant, and needs to be measured for each spring. Note that the direction of Fs is always opposite to the direction of 13

FA = kx

x

the displacement x. That is why we have the minus sign in Hooke’s Law. It is a restoring force. We find that the potential energy of the compressed or stretched spring, measured from its equilibrium position is the same as the work done by the applied force. This is the same as the area under the triangle in the figure above and can be written as, 1 P Es = kx2 2

6-5 Conservative and Nonconservative Forces If friction is present, the work done depends not only on the starting and ending points, but also on the path taken. Friction is called a nonconservative force. Below you can see the work done by friction is different for the two paths going from a to b.

path 1

Ffr

F

fr

Ffr path 2 a

Ffr

b

14

Potential energy can only be defined for conservative forces. Therefore, we distinguish between the work done by conservative forces and the work done by nonconservative forces. Work done by a conservative force such as gravity or elastic can be written as: Wg = −mgy = −∆P Eg 1 Ws = − kx2 = −∆P Es 2 WC = −∆P E In general. Note the works done are negatives of the corresponding PEs. The total or net work done is the sum of conservative and nonconservative forces: Wnet = WC + WN C = −∆P E + WN C = ∆KE since Wnet = ∆KE WN C = ∆KE + ∆P E We find that the work done by nonconservative forces is equal to the total change in kinetic and potential energies.

6-6 Mechanical Energy and Its Conservation If there are no nonconservative forces, the sum of the changes in the kinetic energy and in the potential energy is zero - the kinetic and potential energy changes are equal but opposite in sign. ∆KE + ∆P E = 0 (KE2 − KE1 ) + (P E2 − P E1 ) = 0 KE2 + P E2 = KE1 + P E1

This allows us to define the total mechanical energy: E = KE + P E E2 = E1 conservation of mechanical energy 15

6-7 Problem Solving Using Conservation of Mechanical Energy Consider a mass m released from rest at point y1 . With what velocity does it hit the spring and how much did the spring compressed? y

PEg

1

KE + PEg

y

v2 = ?

2

KE + PEg + PE y

3

y=0

S

PES

Conservation of energy gives, E1 = E2 = E3 KE1 + P E1 = KE2 + P E2 1 0 + mgy1 = mv22 + mgy2 2 p v2 = 2g(y1 − y2 ) KE1 + P E1 = KE3 + P E3 1 0 + mgy1 = k(y2 − y3 )2 2 r 2gy1 y2 − y3 = k (6-40) A roller-coaster car shown in the figure is pulled up to point 1 where it is released from rest. Assuming no friction, calculate the speed at points 2, 3, and 4. 16

1 3 32m 4 26m

14m

2

Solution Since there are no dissipative forces present, the mechanical energy of the roller coaster will be conserved. Subscript 1 represents the coaster at point 1, etc. The height of point 2 is the zero location for gravitational potential energy. We have v1 = 0 and y1 = 32 m. Point 2: ⇒ Point 3: ⇒ Point 4: ⇒

1 1 1 mv12 + mgy1 = mv22 + mgy2 ; y2 = 0 ⇒ mgy1 = mv22 2 2 2 p v2 = 2gy1 = 25 m/s. 1 1 1 mv2 + mgy1 = mv32 + mgy3 ; y3 = 26 m ⇒ mgy1 = mv32 + mgy3 2 2 2 1p v3 = 2g(y1 − y3 ) = 11 m/s. 1 1 1 mv2 + mgy1 = mv42 + mgy4 ; y4 = 14 m ⇒ mgy1 = mv42 + mgy4 2 2 2 1p v4 = 2g(y1 − y4 ) = 19 m/s.

(6-41) Chris jumps off a bridge with a bungee cord (a heavy stretchable cord) tied around his ankle, Fig. 6-42. He falls for 15 m before the bungee cord begins to stretch. Chris’s mass is 75 kg and we assume the cord obeys Hooke’s law, F = –kx, with k = 55 N/m. If we neglect air resistance, estimate what distance d below the bridge Chris’s foot will be before coming to a stop. Ignore the mass of the cord (not realistic, however) and treat Chris as a particle. Solution The only forces acting on the bungee jumper are gravity and the elastic force from the bungee cord, so the jumper’s mechanical energy is conserved. Subscript 1 represents the jumper at the bridge, and subscript 2 represents 17

the jumper at the bottom of the jump. Let the lowest point of the jumper’s motion be the zero location for gravitational potential energy ( y = 0). The zero location for elastic potential energy is the point at which the bungee cord begins to stretch. See the diagram in the textbook. We have v1 = v2 = 0, y1 = d, y2 = 0, and the amount of stretch of the cord x2 = d − 15. Solve for d. Note that we ignore Chris’s height, since his center of mass falls farther than d. 1 1 1 1 E1 = E2 ⇒ mv12 + kx12 + mgy1 = mv22 + kx22 + mgy2 2 2 2 2 i h 1 mg mgd = k(d − 15)2 ⇒ d2 − 30 + 2 + 225 = 0 k√ 2 56.7 ± 56.72 − 900 ⇒ d2 − 56.7d + 225 = 0 ⇒ d = 2 d = 52.4, m 4.29 therefore d = 52.4 m. (6-44) A block of mass m is attached to the end of a spring (spring stiffness constant k), Fig. 6-43. The mass is given an initial displacement x0 from equilibrium, and an initial speed v0 . Ignoring friction and the mass of the spring, use energy methods to find (a) its maximum speed, and (b) its maximum stretch from equilibrium, in terms of the given quantities. Solution At the release point the mass has both kinetic energy and elastic potential energy. The total energy is 12 mv02 + 12 kx02. (a) The mass has its maximum speed at a displacement of 0, so it has only kinetic energy at that point. r 1 2 1 2 k 1 2 mv0 + kx 0 = mvmax ⇒ vmax = v 20 + x20 2 2 2 m (b) The mass has a speed of 0 at its maximum stretch from equilibrium, so it has only potential energy at that point. r 1 2 1 2 m 1 2 ⇒ xmax = x20 + v02 mv + kx = mv 2 0 2 0 2 max k (6-39) A vertical spring (ignore its mass), whose spring constant is 875 N/m, is attached to a table and is compressed down by 0.160 m. Solution 18

v3 = 0 v2 x=0 y=0

v1 = 0

1

2

3

Use conservation of energy. The level of the ball on the uncompressed spring is taken as the zero location for both gravitational potential energy (y = 0) and elastic potential energy (x = 0). (a) What upward speed can it give to a 0.380-kg ball when released? Subscript 1 represents the ball at the launch point, and subscript 2 represents the ball at the location where it just leaves the spring, at the uncompressed length. We have v1 = 0, x1 = y1 = −1.6 m, and x2 = y2 = 0. E1 = E2 1 1 1 1 ⇒ mv21 + kx21 + mgy1 = mv22 + kx22 + mgy2 2 2 2 2 1 2 1 2 ⇒ 0 + kx1 + mgy1 = mv2 + 0 + 0 2 2 solve for v2 r kx21 + 2mgy1 = 7.47 m/s. ⇒ v2 = m (b) How high above its original position (spring compressed) will the ball fly? Subscript 3 represents the ball at its highest point. We have v1 = 0, x1 = y1 = −1.6 m, and x3 = 0, v3 = 0.

19

Solve for y3 . E1 = E3 1 1 1 1 ⇒ mv21 + kx21 + mgy1 = mv32 + kx23 + mgy3 2 2 2 2 1 2 ⇒ 0 + kx1 + mgy1 = 0 + 0 + mgy3 2 kx21 = 3.01 m. ⇒ y3 − y1 = 2mg (6-90) A ball is attached to a horizontal cord of length l whose other end is fixed, l 1 3

h

Peg

r 2

(a) If the ball is released, what will be its speed at the lowest point of its path? Solution: The tension in the cord is perpendicular to the path at all times, so the tension in the cord doesn’t do work on the ball. Only gravity does work on the ball, so the mechanical energy of the ball is conserved. Subscript 1 represents the ball when it is horizontal, and subscript ...