1. Physics 1501: Lecture 13 Announcements HW 05: Friday next week
Midterm 1: Monday Oct. 3
Covers Chapters 1-5
Practice test + solutions on web
HW solutions: on the web.
Office hours today
Topics
Review Work & Energy, Power
Potential energy
Conservative & non-conservative forces

2. Definition of Kinetic Energy : Work Kinetic-Energy Theorem:
Review
Definition of Work:
Work (W) of a constant force F acting through a displacement  r is:
W = F .  r = F  r cos  = Fr  r
Definition of Kinetic Energy :
The kinetic energy of an object of mass (m) moving at speed (v) is:
K = 1/2 m v2
Work Kinetic-Energy Theorem:

3. Work & Power:
Consider the following,
But,
So, Z3
GLC

4. Work Done Against Gravity
Consider lifting a box onto the tail gate of a truck. m h
The work required for this task is,
W = F · d
= (mg)(h) (1)
W = mgh

5. Work Done Against GravityF m
mgsinq
mgcosq h q mg
To push the box with constant speed, F = mgsinq
The length of the ramp is h/sinq
So the work done is,
W = Fd
= (mgsinq)(h/sinq)
W = mgh
Same as before !

6. Work Done by a Spring Fs Dx Force from the spring is Fs = -kx,
Displacement is x.
W =  F dx
= - kx dx
W = - 1/2 kx2
Remember these two results for Chapter 7.

7. Work and Power
What is the power required for a car (m=1000 kg) to climb a hill (5%) at v=30m/s assuming the coefficient of friction m = 0.03 ?
V = 30 m/s
Car 5%

8. Solution m g  The car needs to spend that power to get up the hill j
FBD
P =  F·v
= f·v + m g·v 5%
m g f N
v=30 m/s 
f = -|N| i = - mg cos  i
g = - g sin  i - g cos  j
P = -  mg cos  v - mg sin  v
= - mg v (  cos  + sin  )
= - mg v (0.03  )/( )1/2
= - (1000 kg)(10 m/s2)(30 m/s)(0.08)
= W ~ -24 kW 5
( )1/2
100
The car needs to spend that power to get up the hill

9. Chap.7: New Topic - Potential Energy
Consider a ball at some height above the ground.
No Velocity
Some Velocity

10. New Topic - Potential Energy
Consider a ball at some height above the ground.
What work is done in this process ?
(Work done by the earth on the ball)
W = F. Dx
W = mgh cos(0)
W = mgh h

11. New Topic - Potential Energy
Consider a ball at some height above the ground.
Before the ball falls it has the potential to do an amount of work mgh.
We say the ball has a potential energy of U = mgh.
By falling the ball loses its potential energy, work is done on the ball, and it gains some kinetic energy,
W = K = 1/2 mv2 = -DU = mgh h

12. New Topic - Potential Energy
What if the path it follows to the ground is different ?
h/2 h 45
h/2 45

13. New Topic - Potential Energy
What if the path it follows to the ground is different ? mg q
mgsinq
mgcosq y
Force of gravity F1 = + mgsinq
Distance is d1=(+h/2)/sinq
W1 = F1 d1
= mgh/2 h 45
h/2
F2 = - mgsinq
d2 =(-h/2)/sinq
W2 = (-mgsinq)(-h/2sinq)
= mgh/2
W = W1 + W2 = mgh

14. New Topic - Potential Energy
How much work is done when the ball has fallen halfway ? h
h/2

15. Lecture 13, ACT 1 Work Done by Gravity
The air track is at an angle of 30 degrees with respect to horizontal. The cart (with mass 1 kg) is released. It bounces back and forth on the track. It falls 1 meter down the track, then bounces back up to its original position. How much total work is done by gravity on the cart when it reaches its original position.
1 meter
30 degrees
A) 5 J B) 10 J C) 20 J D) 0 J

16. Lecture 13, ACT 1 Solution d = 1 m mg d = 1 m mg
Do this problem in two steps, going down and going up
Going down:
Wd = F.Dx
Wd = [-mg sin(30)] [-d] =mgd/2
30 degrees
d = 1 m mg
Going up:
Wu = F.Dx
Wu = [-mg sin(30)] [d] = - mgd/2
30 degrees
d = 1 m mg
Total:
W = Wd + Wu
W = mgd ( 1/2 – 1/2 ) = 0
The answer is D) 0 J
Note: W didn’t depend on path

17. Some Definitions
Conservative Forces - those forces for which the work done does not depend on the path taken, but only the initial and final position.
Potential Energy - describes the amount of work that can potentially be done by one object on another under the influence of a conservative force
W = -DU
only differences in potential energy matter.

18. Potential Energy
For any conservative force F we can define a potential energy function U in the following way:
The work done by a conservative force is equal and opposite to the change in the potential energy function.
This can be written as:
W = F.dr = - U ò r1 r2 U2 U1
U = U2 - U1 = - W = - F.dr ò r1 r2

19. A Conservative Force : Spring
For a spring we know that Fx = -kx.
F(x) x1 x2 x
relaxed position
-kx
F = - k x1
F = - k x2

20. What is the Work done by the Spring...
The work done by the spring Ws during a displacement from x1 to x2 is the area under the F(x) vs x plot between x1 and x2. x2 x1
F(x) x Ws
kx1
kx2
-kx

21. A Non-Conservative Force Friction
Looking down on an air-hockey table with no air,
Path 2
Path 1
For which path does friction do more work ?

22. A Non-Conservative Force
Path 2
Path 1
Since |W2|>|W1| the puck will be traveling slower at the end of path 2.
Work done by a non-conservative force takes energy out of the system.
W1 = -mmg d1
W2 = -mmg d2
since d2 > d1, -W2 > -W1

23. Lecture 13, ACT 2 Work/Energy for Conservative Forces
The air track is again at an angle of 30 degrees with respect to horizontal. The cart (with mass 1 kg) is released 1 meter from the bottom and hits the bumper with some speed, v1. You want the cart to go faster, so you release the cart higher. How high do you have to release the cart so it hits the bumper with speed v2 = 2v1?
1 meter
30 degrees
A) 1 m B) 2 m C) 4 m D) 8 m

24. Lecture 13, ACT 2 Work / Energy
Case 1
Work done W = mgL1/2;
(Note DU = -mgDy = -mgh1)
Work energy theorem,
W = DK
1/2 mgL1 = 1/2 m(v12-0)
L1 = v12/g
Case 2
Work done W = mgL2/2;
(Note DU = -mgDy = -mgh2)
Work energy theorem,
W = DK
1/2 mgL2 = 1/2 m(v22-0)
1/2 mgL2 = 1/2 m((2v1)2-0)
L2 = 4v12/g = 4h1
1 meter
30 degrees
C) 4 m

25. Lecture 13, ACT 3 Work/Energy for Non-Conservative Forces
The air track is once again at an angle of 30 degrees with respect to horizontal. The cart (with mass 1 kg) is released 1 meter from the bottom and hits the bumper with some speed, v1. This time the vacuum/ air generator breaks half-way through and the air stops. The cart only bounces up half as high as where it started. How much work did friction do on the cart ?
1 meter
30 degrees
A) 2.5 J B) 5 J C) 10 J D) –2.5 J E) –5 J F) –10 J

26. Lecture 13, ACT 3 Work / Energy for Non-Conservative Force
Start is at top, v=0. Finish is half way up when v=0.
Velocity starts and ends at zero.
Total Work done on the cart must be zero.
Work done by gravity Wg = mgDy = mgL/4;
Work done by friction must cancel this,
WT = Wg + Wf = 0
Wf = - Wg
Wf = -mgL/4 = -(1kg)(10m/s2)(1m)/4 = -2.5J
1 meter
30 degrees
D) -2.5 J

27. Conservation of Energy
If only conservative forces are present, the total energy (sum of potential and kinetic energies) of a system is conserved.
E = K + U
E = K + U
= W + U
= W + (-W) = 0
using K = W
using U = -W
E = K + U is constant !!!
Both K and U can change, but E = K + U remains constant.