201: Physics Lecture 10
Work & Energy Definition Work done by a constant force Work/kinetic energy theorem Work done by multiple constant forces
3/27/08
Physics 201, UW-Madison
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Definition of Work:
Ingredients: Force (F), displacement (r) Work, W, of a constant force F acting through a displacement r is:
F Fr r
nt e
! ! W = Fi!r = F!r cos" = Fr !r
Vector "Dot Product"
d
p la is
em c
3/27/08
Physics 201, UW-Madison
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Definition of Work...
Only the component of F along the displacement is doing work. Example: Train on a track. If you push on it at =90o the train does not move; no work is done.
F r F cos
3/27/08
Physics 201, UW-Madison
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Question
You lift an object 1 m from the floor to a shelf. The work done by the earths gravitational field on the object is: Positive No work is done Negative Insufficient information
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Physics 201, UW-Madison
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Question
You push a heavy box 1 m across a rough floor. Which statements are correct? You have done positive work on the box. You have done negative work on the box. The earths gravitational field has done negative work on the box The normal force has done positive work on the box You have done negative work on the outside world
3/27/08
Physics 201, UW-Madison
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Back to the definition of Work:
Total Work, W, of a force F acting through a displacement x is:
Fx
W = " ( Fx !x )i
i
x2
Wtot =
x1
! F dx
x
Fi x xi
Work = Area under F(x) curve
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Problem 1: Work & Energy
A box is pulled up a rough ( > 0) incline by a rope-pulleyweight arrangement as shown below. How many forces are doing work on the box?
(a) 2 (b) 3 (c) 4
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Physics 201, UW-Madison
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Problem 1: Solution
Draw FBD of box:
T
v
N
Consider direction of motion of the box Any force not perpendicular to the motion will do work: N does no work (perp. to v) T does positive work f does negative work 3 forces do work mg f
mg does negative work
3/27/08
Physics 201, UW-Madison
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Work: 1-D Example (constant force)
A force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m.
F
x Work done by F on box : WF = F
. x = F x
(since F is parallel to x)
WF = (10 N) x (5 m) = 50 Joules (J)
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Work & Kinetic Energy:
A force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m. The speed of the box is v1 before the push and v2 after the push.
v1 F
v2 m i x
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Physics 201, UW-Madison
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Work & Kinetic Energy...
Since the force F is constant, acceleration a will be constant. We have shown that for constant a: v22 - v12 = 2a(x2-x1) = 2ax. 1/ mv 2 - 1/ mv 2 = max multiply by 1/2m: 2 2 2 1 1/ mv 2 - 1/ mv 2 = Fx But F = ma 2 2 2 1
v1 F
v2 m a i x
3/27/08
Physics 201, UW-Madison
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Work & Kinetic Energy...
So we find that 1/2mv22 - 1/2mv12 = Fx = WF Define Kinetic Energy K: K = 1/2mv2 K2 - K1 = WF WF = K (Work/kinetic energy theorem) v1 F v2 m a i x
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Physics 201, UW-Madison
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Problem 2: Work & Energy
Two blocks have masses m1 and m2, where m1 > m2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e. > 0) which slows them down to a stop. Which one will go farther before stopping?
(a) m1 (b) m2
(c) they will go the same distance
m1
m2
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Problem 2: Solution
The work-energy theorem says that for any object WNET = K In this the example only force that does work is friction (since both N and mg are perpendicular to the block's motion).
N f
m mg
3/27/08
Physics 201, UW-Madison
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Problem 2: Solution
The work-energy theorem says that for any object WNET = K In this example the only force that does work is friction (since both N and mg are perpendicular to the blocks motion). The net work done to stop the box is - f D = -mgD.
This work "removes" the kinetic energy that the box had: WNET = K2 - K1 = 0 - K1
m
D
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Problem 2: Solution
The net work done to stop a box is - f D = -mgD. This work "removes" the kinetic energy that the box had: WNET = K2 - K1 = 0 - K1 This is the same for both boxes (same starting kinetic energy).
m2gD2 = m1gD1 Since m1 > m2 we can see that
m2D2 = m1D1 D2 > D1
m1 D1
3/27/08 Physics 201, UW-Madison
m2 D2
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A falling object
What is the speed of an object after falling a distance H, assuming it starts at rest? Wg = F r = mg r cos(0) = mgH
.
v0 = 0 r mg j
Wg = mgH Work/Kinetic Energy Theorem: Wg = mgH = 1/2mv2
H
v = 2gH
3/27/08 Physics 201, UW-Madison
v
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What about multiple forces?
Suppose FNET = F1 + F2 and the displacement is r. The work done by each force is: W 1 = F 1 r
.
W2 = F 2 r
.
F1
FNET
r
WTOT = W1 + W2 = F1 r + F2 r = (F 1 + F 2 ) r WTOT = FTOT r
.
.
F2
.
.
It's the total force that matters!!
3/27/08
Physics 201, UW-Madison
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Comments:
Time interval not relevant Run up the stairs quickly or slowly...same W
Since W = F r
.
No work is done if: F=0 or r = 0 or = 90o
3/27/08
Physics 201, UW-Madison
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Comments...
W = F r
.
No work done if = 90o. No work done by T.
T v
v
No work done by N.
N
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Physics 201, UW-Madison
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Question
When you throw a 0.2 kg ball upwards at 45 degrees, it rises to 6m maximum height above you. Which statements are correct concerning the kinetic energy at the maximum height?
A. The kinetic energy is zero B. The kinetic energy is unchanged C. The kinetic energy decreases by half D. The kinetic energy decreases a small amount E. Insufficient information
Horizontal component of velocity, vcos45, remains unchanged - kinetic energy decreases by the square of 1/cos45, i.e., half.
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Question
From the edge of a cliff, you throw the same ball downwards at 45 degrees to the horizontal with the same initial velocity. What is the velocity when it hits the ground below the cliff?
A. 7.7 m/s B. 10.8 m/s C. 15.3 m/s D. Insufficient information
Height of the cliff from the ground below is not stated.
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Physics 201, UW-Madison
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Question
The work needed to maintain uniform circular motion in a horizontal plane is:
A. zero B. Positive and constant C. Negative and constant D. Positive and variable E. Negative and variable
Force applied (towards the center of the circle) is perpendicular to the direction of motion at any point in the UCM.
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Question
A 5 kg mass is hanging quietly from a spring. The spring has elongated by 10 cm. How much work is the spring now doing?
A. 0.49 J B. 0 C. -0.49 J
As long as the mass is not moving no work is getting done.
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Physics 201, UW-Madison
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Question
Since the kinetic energy is always positive, the work done must always be positive as well.
A. False B. True
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Physics 201, UW-Madison
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Question
The static frictional force can never do work.
A. False B. True
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Physics 201, UW-Madison
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