1
New policy for workshops
You are expected to print, read, and think about the workshop material prior to coming to class. (This part of the policy is not new!) There will be a “prelab” question which is to be answered on the worksheet before coming to class. The prelab question will be graded during the quiz. (This policy replaces the penalty for not printing the worksheet in advance.)
Work & Kinetic Energy
Review of Previous Lecture
• Fluid resistance, terminal speed
at high speed
• Uniform circular motion:
• If net force on object is perpendicular to v:
• Net force causes radial acceleration
via Newton’s 2nd law.
• Object’s path becomes circular with
radius R:
Tangent to circle ; v = const
Radially inward ; a = const
v
Dmgvmgf
Dvf
/
2
v
a
var
2
Rv
ra
RmvF
2
a
vR
rF
v
Rm
• The net force pointing to the center is called “radial” or
“centripetal” force.
• The centripetal force may be made up of weight,
friction, tension, normal forces acting on object – it is
not an additional force!
A block is given a kick and then freely slides up an inclined plane at angle q with the horizontal. There is friction between the block and plane. The magnitude of the acceleration of the block is
A. g
B. g sinq
C. g cosq
D. larger than g sinq
E. smaller than g sinq
i-Clicker
4
i-Clicker
A B C
A soccer ball is dropped from a hot air balloon. During the fall, air resistance cannot be neglected. Which of the following graphs best represents the acceleration vs. time graph? (Choose down direction as positive.)
A ball is attached to the end of a cord of length 2 meters. The ball is whirled in a vertical circle. Which is correct?
A. The tension at the bottom and at the top is greater than the weight of the ball.
B.The tension at the bottom and the top is less than the weight of the ball.
C.The tension at the bottom is greater, and at the top less than the weight.
D. The tension at the bottom is greater than the weight, and at the top it may be greater or less than the weight.
E. The tension at either location may be greater or less than the weight, depending on the speed.
i-Clicker
A 2 kg ball is attached to the end of a cord of length 2 meters. The ball is whirled in a vertical circle. The tension in the cord at the top of the circle is 0 N. Which is correct?
A. The acceleration at the top of the circle is zero.
B. The acceleration at the top of the circle is 9.8 m/s2, upward.
C. The acceleration at the top of the circle is 9.8 m/s2 downward.
D.The acceleration at the top of the circle has magnitude 19.6 m/s2.
E. The acceleration at the top of the circle cannot be determined.
i-Clicker
VECTOR MULTIPLICATION:
SCALAR PRODUCT
Vector Addition:
What does mean?
Scalar Product (= Scalar)
(projection of on ):
Note:
Scalar product of two vectors is COMMUTATIVE.
BA
BA
BA
BA
qcosABBA
ABA
B
A
BAAB
q
coscos
cos
ABBA
ABAB
A
B
A B
A
B
BA
BA
A
B
qcosB
q
qcosA
qB
A
SCALAR PRODUCT USING COMPONENTS
Dot product is distributive
Consider:
Use dot product to find angle between vectors.
Example:
)( CBA
CABA
jAiAA yxˆˆ
jBiBB yxˆˆ
)ˆˆ()ˆˆ( jBiBjAiABA yxyx
jjBAijBAjiBAiiBA yyxyyxxxˆˆ ˆˆ ˆˆ ˆˆ
)( zzyyxx BABABA
j
1ˆˆˆˆ jjii
0ˆˆ ji
2|| ; ˆˆ AkiA
2|| ; ˆˆ BkjB
1)ˆ()ˆˆ( kjkiBA
cos|||| BABA
!60 cos2
1
||||cos
211 o
BA
BA
A
B
BA
CA
C
CB
A
B
1 0 0 1
i
J )3)(cos5(
ˆ)m3(]ˆ)sinN5(ˆ)cosN5[(
ˆ)inN5(ˆ)cosN5(
q
ijiW
jsiF
WORK
Suppose block of mass m lies on horizontal plane.
Case A: A horizontal
force acts continuously
as block is displaced by
Work done by force:
Case B: Force and
displacement are
not collinear
Work = (displacement) X (component of F along displacement)
EXAMPLE:
F
s
WsFsFW
)cos( q
J 15 Nm 15]ˆ)m3[(]ˆ)N5[(
ˆ)m3( , ˆ)N5(Let
iiW
isiF
F
m
s
F
m
s
q
A:
B:
FsW
Work done (by a force) may be:
• Positive
• Negative
• Zero
EXAMPLES:
• Positive: e.g. block on previous viewgraph
• Negative:
Arnold lowers
the barbell
• Zero: Uniform circular motion
tangent ||
radial
vsd
Fc
F
s
v
CF
n – mgcosq = 0
n = mgcosq
mgsinq – fK
fK = mKn = mKmgcosq
qmqm cosˆˆcos mgsisimgsfW KKKf
jmgimgwFGravˆcosˆsin qq
qsinmgsswWg
EXAMPLE:
A block of mass m slides
a distance s down an incline
plane.
How much work is done
by friction?
How much work is done by gravity?
i
yi
F
i
xi
F
s
q
m
m
n
q mgcosq
mgsinq
fK
mg
What is the net work done by all forces acting on the
block as it slides?
jmgnimg
jFiFFF
K
i i i
yxinet ii
ˆ)cos(ˆ)cos(sin
ˆ)(ˆ)(
qqmq
qmq cossin Kx mgssF SFW netnet
A block is pushed so that it moves distance L up a ramp (incline angle q ) at constant speed. If there is friction, the work done on the block by the hand…
A. is mg sinq L.
B. is less than mg sinq L.
C. is greater than mg sinq L.
D. could be greater of less.
E. is zero.
i-Clicker
Is the work positive or negative?
A block is pushed so that it moves distance L up a ramp (incline angle q ) at constant speed. The work done on the block by the normal force of the ramp…
A. is mg cosq L.
B. is less than mg cosq L (but > 0)
C. is greater than mg cosq L.
D. could be greater of less.
E. is zero.
i-Clicker
A block is pushed so that it moves distance L up a ramp (incline angle q ) at constant speed. If there is friction, the magnitude of the work done on the block by friction…
A. is mg sinq L.
B. is less than mg sinq L.
C. is greater than mg sinq L.
D. could be greater of less than mg
sinq L.
E. is zero.
i-Clicker
Is the work positive or negative?
A block is pushed so that it moves distance L up a ramp (incline angle q ) at constant speed. If there is friction, the magnitude of the work done on the block by gravity…
A. is mg sinq L.
B. is less than mg sinq L.
C. is greater than mg sinq L.
D. could be greater of less than mg
sinq L.
E. is zero.
i-Clicker
Is the work positive or negative?
A block is pushed so that it moves distance L up a ramp (incline angle q ) at constant speed. If there is an unknown amount of friction, the net work done on the block…
A. is zero.
B. is less than zero.
C. is greater than zero.
D. could be positive or negative.
E. need to know the friction force.
i-Clicker
2
1212
221 mvmvWtot
2
21 mvK
Work done by a net force on an object of mass m = change in kinetic energy of the object
WORK AND KINETIC ENERGY
Consider collinear force and displacement:
Where Fnet is NET force
RECALL: If constant acceleration a
changes v1 to v2 over a distance s
Define Kinetic Energy:
Work-Energy Theorem:
sFW nettot
a
vvs
2
2
1
2
2
2
2
1
2
2 vvmsma
Fnet
2
1212
221 mvmvKsFW nettot
m 30
m/s 30kg 10002
21
netF
m
/sm kg
30
900500 22
2
5
s
m kg
30
105.4
N 105.1 4
24
m/s 15kg
N
1000
105.1
a
m
Fnet
What is the corresponding acceleration???
EXAMPLE:
Recall the yellow light
Car with Mass of
1,000 Kg moving at
30 m/s must stop in 30 m.
How much force must be applied?
jsjmgFnetˆ)m 320( ;)ˆ(
2
1212
221)m 320( mvmvmgWtot
)m 320(22
2 gv
/sm 6400 22
m/s 802 v
EXAMPLE:
Drop a bowling ball from the top
of the Empire State Building.
How fast is it going just before it
hits the ground?
Use Work-Energy Theorem:
KSFW nettot
m 320
Two identical blocks are released from rest at the same
height on two separate frictionless ramps. The blocks
reach the same final height labeled finish. Which
statement is correct regarding the work done on the
two blocks by the gravitational force?
A. Block B travels further, so more work is done on block B by the gravitational force than on block A.
B. Both blocks fall the same vertical distance, so the work done by gravity is the same.
C. By Newton’s Third Law, the force exerted on the block by the earth is exactly cancelled by the force exerted on the earth by the block. The work done is zero.
D. The angle between force and displacement is smaller for block A than for block B, thus the work done on block A is greater.
E. None of the above.
i-Clicker
Two identical blocks are released from rest at the same
height at the same time. Block A slides down a steeper
ramp than Block B. Both ramps are frictionless.
The speed of Block A as it crosses the finish line is:
A. greater than the speed of Block B at the finish line.
B. less than the speed of Block B at the finish line.
C. equal to the speed of Block B at the finish line.
D. Not enough information.
i-Clicker
i
iix
tot xxFW )(lim0
2
1
)(x
xtot dxxFW
WORK DONE BY VARYING FORCE
• Consider varying force along line
For constant force:
For varying force:
EXAMPLE: Force to stretch spring
W = Fs
W = Area under
Force vs. displacement
curve
k = spring constant
[N/m]
“Hooke’s Law”
kxFx
Fx
x1 x2
x s
x 2x
F
2F
Fx
x
F(x)
x1 x2
xi
F(xi)
F
2
1
2
1
2
21
x
x
x
xkxdxkx
)( 2
1
2
221 xxk
Work done by F to stretch spring:
In General: Work done on spring to extend from x1 to x2:
NOTE: Work-Energy theorem valid for varying
forces on curved paths!
2
1
)(x
xx dxxFW
2
21
0
2
21 kXkxW
X
X
o
X
ox
kxdx
dxxFW
)(Fx
x
F(x)
X
kX
10 cm
)( 2
1
2
221 xxk
Wtot = 4.00 J
k = 800 N/m
Initial extension:
Additional extension:
22
21 )cm 10()cm 20( kW
22
21 m 010.0m 040.0)N/m 800(
= 12.0 J
EXAMPLE: It takes 4.00J to stretch a spring from
equilibrium to 10.0 cm. How much work must be
done to stretch it an additional 10 cm?
22
21 )m 0.0()m 10.0( k
27
New policy for workshops
You are expected to print, read, and think about the workshop material prior to coming to class. (This part of the policy is not new!) There will be a “prelab” question which is to be answered on the worksheet before coming to class. The prelab question will be graded during the quiz. (This policy replaces the penalty for not printing the worksheet in advance.)