8
Potential Chapter Energy and Conservation of Energy
In this chapter we will introduce the following concepts: Potential Energy Conservative and non-conservative forces Mechanical Energy Conservation of Mechanical Energy The conservation of energy theorem will be used to solve a variety of problems As was done in Chapter 7 we use scalars such as work, kinetic energy, and mechanical energy rather than vectors. Therefore the approach is mathematically simpler.
(8-1)
B h
g
Work and Potential Energy: A tomato of mass m is taken together with the earth as the system we study.
o
o
v
A
v
The tomato is thrown upwards with initial speed vo at point A. Under the gravitational force, it slows down and stops at point B. Then it falls back to point A and reaches the original speed vo.
During the trip from A to B the gravitational force Fg does work
U = W
(8-2)
S
A
o
m
B Fk
Consider our system with the mass m attached to a spring of spring constant k. The mass has an initial speed vo at point A. Under the spring force FS m slows down and stops at point B (with a spring compression x). Then the mass reverses the direction of its motion and reaches point A. Its speed reaches the original value vo.
v
S
o
A F
m
B k
v
During the trip from A to B the spring force Fs does work
U = W
(8-3)
Conservative and non-conservative forces. The gravitational force and the spring force are d A B called conservative because they can vo m transfer energy between kinetic energy and m x potential energy with no loss. Frictional and drag forces are called non-conservative for reasons that are explained below. Consider a block of mass m on the horizontal floor. The block starts to move with initial speed vo at point A. The coefficient of kinetic friction between the floor and the block is k. The block will slow down and stop at point B after traveling a distance d. During the trip from A to B the frictional force does work
f f
(8-4)
(8-5)
Path Independence of Conservative Forces How to decide whether a force is conservative or non-conservative?
Closed-loop test: A force is conservative if the net work done on a particle during a round trip is always equal to zero (see fig.b). W =0
net
In the examples of the tomato-earth and mass-spring system Wnet = Wa1b + Wb2a = 0 (because U = W Wa1b = U a U b , Wb 2 a = U b U a ) So gravitational force and spring force are conservative forces! Next, we prove the path independence of work done by a conservative force
Wa1b = Wa 2b = U a U b
Thus, the work done by a conservative force does not depend on the path taken but only on the initial and final points choose a simple path
Path-independence of work done by conservative forces
Work done by conservative forces (W = - U) is path-independent, it results in changes in the object's potential energy. Let's use gravity an example of a CONSERVATIVE FORCE. The block's final change in potential energy is the same whether it follows the path with the intermediate stops B, C and D or whether it is directly taken from A to E.
Path-dependence of work done by nonconservative forces
The work done by non-conservative forces does depend on the path. For example, a kinetic friction force is exerted on a sliding box moving from position 1 to 2 following two paths. Because Wf= -f d (assuming constant coefficient of kinetic friction) longer path requires more work.
U = W
Determining Potential Energy Values : . x How to determine the difference in potential ..x O energy U of a conservative force F between points x f and xi ? A conservative force F moves an object along the x-axis from an initial point xi
F(x)
to a final point x f . The work W that the force F does on the object is given by :
xf
W=
F ( x)dx
xi
The corresponding change in potential energy U is: U = W Therefore the expression for U becomes:
xf
U = W = F ( x)dx
xi
(8-6)
y
.
m y
Gravitational Potential energy : Consider a particle of mass m moving vertically along the y -axis from yi to y f .
dy mg
. .O
The gravitational force does work W on the particle which changes the potential energy U of the particle-earth system.
U = F ( y )dy
yi
yf
yf
F = mg
yf y
U = ( mg ) dy = mg dy = mg [ y ] y f = mg ( y f yi ) = mg y.
Since only changes in potential energy are physically menaingful, we set U i = 0 we assign y f = y, yi = 0 U ( y ) = mgy
yi
yi
i
U ( y ) = mgy
y
(8-7)
Potential Energy of a spring :
x O (a) x O (b) x O (c)
Consider the block-mass system. The block moves from xi to x f . The spring force does work W on the block which changes the potential energy of the block-spring system
kxi2 U = W = F ( x)dx = kxdx = k xdx = 2 2 xi xi xi
xf xf xf
kx f 2
Since only changes in potential energy are physically meaningful, we set U i = 0 kx 2 we assign x f =x, xi = 0, U ( x) = 2
x
kx 2 U= 2
x
(8-8)
Conservation of Mechanical Energy : Mechanical energy of an object is defined as the sum of potential and kinetic energies Emech = K + U work-kinetic-energy theorem K = Wnet equations 1 and 2 K = U (eqs.1) (eqs.2) If only conservative forces do work Wnet = Wc , U = Wc K f Ki = ( U f U i ) Ki + U i = K f + U f
(8-9)
This equation is known as the principle of conservation of mechanical energy. It can be summarized as: Emech = K + U = 0
If the forces are a mixture of conservative and non-conservative forces Wnet = Wc + Wnc = U + Wnc
Emech = K + U = Wnc
Here, Wnc is the work of all the non - conservative forces of the system
Practice question
8.7.2. A rubber ball is dropped from rest from a height h. The ball bounces off the floor and reaches a height of 2h/3. How can we use the principle of the conservation of mechanical energy to interpret this observation? a) During the collision with the floor, the floor did not push hard enough on the ball for it to reach its original height. b) Some of the balls potential energy was lost in accelerating it toward the floor. c) The force of the earths gravity on the ball prevented it from returning to its original height. d) Work was done on the ball by the gravitational force that reduced the balls kinetic energy. e) Work was done on the ball by non-conservative forces that resulted in the ball having less total mechanical energy after the bounce.
8.7.2. A rubber ball is dropped from rest from a height h. The ball bounces off the floor and reaches a height of 2h/3. How can we use the principle of the conservation of mechanical energy to interpret this observation? a) During the collision with the floor, the floor did not push hard enough on the ball for it to reach its original height. b) Some of the balls potential energy was lost in accelerating it toward the floor. c) The force of the earths gravity on the ball prevented it from returning to its original height. d) Work was done on the ball by the gravitational force that reduced the balls kinetic energy. e) Work was done on the ball by non-conservative forces that resulted in the ball having less total mechanical energy after the bounce.
Emec = mg (2h / 3 h) = mgh / 3 = Wnc
Mechanical Energy Conservation for projectile motion
K= m(v 2 + v 2 ) x y 2 2 + v0 y ) =
2 m(v0x + (voy gt) 2 )
2
gt 2 = mg voy t 2 2 = K 0 mg(y y 0 ) = K 0 (U g U 0 ) K + Ug = K 0 + U0
2 m(v0x
(8-10)
An example of conservation of mechanical energy: a pendulum bob of mass m moving under the gravitational force The total mechanical energy of the bob-earth system is constant.
L h
U = mgh = mgL(1 cos ) K = mv 2 / 2 U max = K max
As the pendulum swings, the energy E is transferred back and forth between kinetic energy K of the bob and gravitational potential energy U of the bob-earth system
When is K at maximum? frame (a), (e) (U is at minimum there). When is U at maximum? frame (c), (g) (K is at minimum there)
Mechanical Energy Conservation during a roller coaster ride
ME=PE1 =50*9.8*4=1960J
More on roller
PE 2 =50*9.8*3=1470J KE 2 = ME PE 2 = 490J v 2 = 2KE 2 / m = 4.43m / s PE 3 =0J KE 3 = ME PE 3 = 1960J v3 = 2KE 3 / m = 8.85m / s KE 4 = mv 2 / 2 = 900J 4 PE 4 = ME KE 4 = 1060J h 4 = PE 4 / mg = 2.16m
Energy Transformation for Downhill Skiing
h d
mv 2 mgh = 2 mv 2 0 = fd = k mgd 2 mgh = k mgd d = h / k
Review of previous lecture
Mechanical energy of an object is defined as the sum of potential and kinetic energies Emech = K + U Change in potential energy is defined in terms of the work done by a conservative force: U = Wc (eqs.2) work-kinetic-energy theorem K = Wnet =Wc +Wnc (eqs.1)
If we compare equations 1 and 2 we have: K + U = Emech = Wnc If Wnc = 0, Emech = 0, which is known as the principle of conservation of m echanical energy.
Example 1
vf
vi d
h
60. A cookie jar is moving up a 40 incline. At a point 55 cm from the bottom of the incline (measured along the incline), the jar has a speed of 1.4 m/s. The coefficient of friction kinetic between jar and incline is 0.15. (a) How much farther up the incline will the jar move? (b) How fast will it be going when it has slid back to the bottom of the incline? (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)?
Example 1
vf
vi
d d
60. A cookie jar is moving up a 40 incline. At a point 55 cm from the bottom of the incline (measured along the incline), the jar has a speed of 1.4 m/s. The coefficient of kinetic friction between jar and incline is 0.15. (a) How much farther up the incline will the jar move? (b) How fast will it be going when it has slid back to the bottom of the incline? (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)?
Finding the Force F ( x) from the potential energy U ( x)
. O
A x
.
F
. x x+
B
x
Consider an object that moves along the x-axis under the influence of an unknown conservative force F given its potential energy U ( x). The object moves from point A (at x) to point B (at x + x). The force F does work W on the object: W = F x (eqs.1) The potential energy U of the system is changed by U = W (eqs.2) If we combine equations 1 and 2 we get: U x We take the limit as x 0 and we get the following equation: F =
dU ( x) F ( x) = dx
U = F ( x)dx
xi
xf
(8-11)
The potential Energy Curve
If we plot the potential energy U versus x for a force F that acts along the x-axis, we can get a lot of information about the motion of a particle on which F is acting.
x2 x3 x4
We can determine the force F(x) using the equation:
F ( x) =
dU ( x) dx
An example is given in the figures above. In fig.a we plot U(x) versus x. In fig.b we plot F(x) versus x.
At x2 , x3 and x4 the slope of the U(x) vs x curve is zero, thus F = 0.
The slope dU/dx between x3 and x4 is negative F > 0 for this interval. The slope dU/dx between x2 and x3 is positive F < 0 for this interval.
(8-13)
Turning Points : The total mechanical energy is Emec = K ( x) + U ( x) Emec is constant (equal to 5 J in the figure) and is thus represented by a horizontal line. We can solve this equation for K ( x):
allowed
x1
forbidden
K ( x) = Emec U ( x)
mv 2 Because kinetic energy K = 0, we can determine which regions 2 of the x-axis motion is allowed: K ( x) = Emec U ( x) 0 If K = Emech U ( x) 0 U ( x) Emec Motion is allow ed If K = Emech U ( x) < 0 U ( x) > Emec Motion is forbidden The points at which Emec = U ( x) are known as turning points for the motion. For example x1 is the turning point for the U versus x plot above. At the turning point K = 0.
(8-14)
Given the U(x) versus x curve the turning points and the regions for which motion is allowed depends on the value of the mechanical energy Emec
1 5 In the above figure, if Emec = 4 J:
x
x
The turning points (Emec = U ) occur at x1 and x > x5. Motion is allowed (Emec U ) for x > x1 If we reduce Emec to 3 J or 1 J the turning points and regions of allowed motion change accordingly. Equilibrium Point: a position at which the slope dU/dx = 0 (and thus F = 0)
stable equilibrium: positions of minima in the U versus x curve unstable equilibrium: positions of maxima in the U versus x curve neutral equilibrium: a region for which F = 0.
For example, if Emec = 4 J, the kinetic energy K = 0 if x > x5 and any particle moving in this region will be stationary
(8-15)
dU ( x) F ( x) = dx
Positions of Stable Equilibrium. An example is point x4 where U has a minimum. If Emec = 1 J then K = 0 at point x4. A particle with Emec = 1 J is stationary at x4. If we displace slightly the particle either to the right or to the left of x4, the force will bring it back to the equilibrium position this equilibrium is stable. Positions of Unstable Equilibrium. An example is point x3 where U has a maximum. If Emec = 3 J then K = 0 at point x3. A particle with Emec = 3 J is stationary at x3. If we displace slightly the particle either to the right or to the left of x3, the force will take it further away from the equilibrium position this equilibrium is unstable
x3 x4
Quiz of Ch8
A small, initially stationary block is released on a frictionless ramp at a height of 3.0 m. Hill heights along the ramp are as shown. The hills have identical circular tops, and the block does not fly off any hill. For which hill is the speed of the block greatest?
D) C) B) A)
A small, initially stationary block is released on a frictionless ramp at a height of 3.0 m. Hill heights along the ramp are as shown. The hills have identical circular tops, and the block does not fly off any hill. For which hill is the speed of the block greatest?
D) C) B) A)
mv 2 mgH = mgh + 2 v = 2 g ( H h)
21. The string in Fig. 8-37 is long, has a ball attached to one end, and is fixed at its other end. The distance d from the fixed end to a fixed peg at point P is 75.0 cm. When the initially stationary ball is released with the string horizontal as shown, it will swing along the dashed arc. What is its speed when it reaches (a) its lowest point and (b) its highest point after the string catches on the peg?
Example 1
(a) vb (b) yb=2r
check mv 2 / r mg b
39. Figure 8-50 shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are , , and . The particle is released at with an initial speed of 7.0 m/s, headed in the negative x direction. (a) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of ? Suppose, instead, the particle is headed in the positive x direction when it is released at at speed 7.0 m/s. (d) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of ?
Example 2
Figure 8-50 shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are , , and . The particle is released at with an initial speed of 7.0 m/s, headed in the negative x direction. (a) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of ? Suppose, instead, the particle is headed in the positive x direction when it is released at at speed 7.0 m/s. (d) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of ?
Example 2
slope=(45-15)/(6-5)=30
>
Figure 8-50 shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are , , and . The particle is released at with an initial speed of 7.0 m/s, headed in the negative x direction. (a) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of ? Suppose, instead, the particle is headed in the positive x direction when it is released at at speed 7.0 m/s. (d) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of ?
Example 2
133. A massless rigid rod of length L has a ball of mass m attached to one end. The other end is pivoted in such a way that the ball will move in a vertical circle. First, assume that there is no friction at the pivot. The system is launched downward from the horizontal position A with initial speed . The ball just barely reaches point D and then stops. (a) Derive an expression for in terms of L, m, and g. (b) What is the tension in the rod when the ball passes through B? (c) A little grit is placed on the pivot to increase the friction there. Then the ball just barely reaches C when launched from A with the same speed as before. What is the decrease in the mechanical energy during this motion? (d) What is the decrease in the mechanical energy by the time the ball finally comes to rest at B after several oscillations?
2 mvo / 2 = mgL 2 mvo / 2 + mgL = mv 2 / 2
Example 3
E=EC-EA=0-
= -mgL
E=EB-EA=0 (
+mgL) = -2mgL
Review of sample test II
y x