Transcript Ppt

Lecture 16
Agenda:

Review for exam

Assignment: For Tuesday, Read chapter 10.1-10.5

M. Tobin’s office hours today are from 3:05 to 3:55 PM
I will have extra office hours tomorrow from 1 to 2:30 PM

Physics 201: Lecture 16, Pg 1
Newton’s Laws
Three blocks are connected on the table as shown.
The table has a coefficient of kinetic friction of 0.350, the masses
are m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg.
m2
m1
a)
T1
m3
If m3 starts from rest how fast is it going after it goes up 2.0 m
Use Work energy theorem DK= WConservative+ WNon-conservative
Physics 201: Lecture 16, Pg 2
Newton’s Laws
The table has a coefficient of kinetic friction of 0.350, the masses
are m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg.
m2
m1
T1
m3
If m3 starts from rest how fast is it going after it goes up 2.0 m
DK=½ m1v1f2 +½ m2v2f2 + ½ m3v3f2 -½ m1v1i2 -½ m2v2i2 - ½ m3v3i2
= ½ (m1+m2+m3) v2
WC. = Wg(mass 1) + Wg (mass 3) = + m1gh - m3gh
WNC = Wfriction = Fx Dx = - m2gm h
½ (m1+m2+m3) v2 = + m1gh - m3gh - m2gm h
v2 = 2gh(m1-m3-m2m)/(m1+m2+m3)
a)
Physics 201: Lecture 16, Pg 3
Work, Energy & Circular Motion

A mass, 11 kg, slides down of a frictionless circular path of
radius, 5.0 m, as shown in the figure. Initially it moves only
vertically and, at the end, only horizontally (1/4 of a circle all
told). Gravity, 10 m/s2, acts along the vertical.
If the initial velocity is 2 m/s downward then
(a) What is the work done by gravity on the mass?
(b) What is the final speed of the mass when it reaches the
bottom?
(c) What is the normal force on the mass
when it reaches the bottom (while still on
the curved sections)?
Physics 201: Lecture 16, Pg 4
Work, Energy & Circular Motion

A mass, 11 kg, slides down of a frictionless circular path of
radius, 5.0 m, as shown in the figure. Initially it moves only
vertically and, at the end, only horizontally (1/4 of a circle all
told). Gravity, 10 m/s2, acts along the vertical.
If the initial velocity is 2 m/s downward then
(a) What is the work done by gravity on the mass?
W = mgR= 11 x 10 x 5 = 550 J
 (b) What is the final speed of the mass
when it reaches the bottom (while still on the curve)?
½ mvf2 = ½ m vi2 + mgR = 22 J + 550 J = 572 J
vf = (1144 / 11) ½ m/s
Physics 201: Lecture 16, Pg 5
Work, Energy & Circular Motion

A mass, 11 kg, slides down of a frictionless circular path of
radius, 5.0 m, as shown in the figure. Initially it moves only
vertically and, at the end, only horizontally (1/4 of a circle all
told). Gravity, 10 m/s2, acts along the vertical.
If the initial velocity is 2 m/s downward then
(c) What is the normal force on the mass
when it reaches the bottom
SFy = m ac = N – mg = m v2 /R
N = mg + m v2 /R = (110 + 11 x 1144/11) N
= 1254 N =1300 N
Physics 201: Lecture 16, Pg 6
Exercise
Work/Energy for Non-Conservative Forces

An air track is at an angle of 30° with respect to horizontal.
The cart (with mass 1.0 kg) is released 1.0 meter from the
bottom and hits the bumper at a 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 ? (g=10 m/s2)
Notice the cart only bounces to a height of 0.25 m
A.
B.
C.
D.
E.
F.
2.5 J
5.0 J
10. J
-2.5 J
-5.0 J
-10. J
30°
h = 1 m sin 30°
= 0.5 m
Physics 201: Lecture 16, Pg 7
Springs



A Hooke’s Law spring with a spring constant of 200 N/m is first
stretched 3.0 m past its equilibrium distance and then is
stretched 6.0 m more meters.
How much work must be done to go from 3.0 m to 9.0 m?
W = Ufinal-Uinitial= ½ k (x-xeq)final2 -½ k (x-xeq)init2
= 100 [(9)2 –(3)2] J= 100(72) J = 7200 J
Physics 201: Lecture 16, Pg 9
Chapter 5 & 6 (Forces and Newton’s Laws)
Physics 201: Lecture 16, Pg 10
Chapter 6
Physics 201: Lecture 16, Pg 11
Chapter 5 & 6
Physics 201: Lecture 16, Pg 12
Chapter 5 & 6
Note: Drag in air is proportional to v or v2
At terminal velocity, drag force cancels out other forces
Physics 201: Lecture 16, Pg 13
Chapter 5 & 6
Physics 201: Lecture 16, Pg 14
Chapter 7 & 8
Physics 201: Lecture 16, Pg 15
Chapter 7 & 8
Physics 201: Lecture 16, Pg 16
Chapter 7 & 8
Note: Ethermal is the same as Einternal (Serway)
Note: Wdissapative is the same as Wnon-conservative
Physics 201: Lecture 16, Pg 17
Chapter 7 & 8
s here refers to
the path
Physics 201: Lecture 16, Pg 18
Chapter 7 & 8
Physics 201: Lecture 16, Pg 19
Work and Energy
A block of mass m is connected by a spring to the ceiling.
The block is held at a position where the spring is
unstretched and then released. When released, the block
(a) remains at rest.

(b) oscillates about the unstretched position
(c) oscillates about a position that is lower than the
unstretched position
(d) oscillates about a position that is higher than the
unstretched position
Physics 201: Lecture 16, Pg 20
Work and Energy
A block of mass m is connected by a spring to the ceiling.
The block is held at a position where the spring is
unstretched and then released. When released, the block
(a) remains at rest.

(b) oscillates about the unstretched position
(c) oscillates about a position that is lower than the
unstretched position
(d) oscillates about a position that is higher than the
unstretched position
Physics 201: Lecture 16, Pg 21
Work and Energy


A mass is attached to a Hooke’s law spring on a horizontal
surface as shown in the diagram below. When the spring is
at its equilibrium length, the block is at position Y.
When released from position X, how will the spring’s
potential energy vary as the block moves from X to Y to Z ?
(a) It will steadily increase from X to Z.
(b) It will steadily decrease from X to Z.
(c) It will increase from X to Y and decrease from Y to Z.
(d) It will decrease from X to Y and increase from Y to Z.
Physics 201: Lecture 16, Pg 22
Work and Energy


A mass is attached to a Hooke’s law spring on a horizontal
surface as shown in the diagram below. When the spring is
at its equilibrium length, the block is at position Y.
When released from position X, how will the spring’s
potential energy vary as the block moves from X to Y to Z ?
(a) It will steadily increase from X to Z.
(b) It will steadily decrease from X to Z.
(c) It will increase from X to Y and decrease from Y to Z.
(d) It will decrease from X to Y and increase from Y to Z.
Physics 201: Lecture 16, Pg 23
Work and Energy

An object moves along a line under the influence of a single
force. The area under the force vs. position graph represents
(a) the impulse delivered to the object
(b) the work done on the object.
(c) the change in the velocity of the object.
(d) the momentum of the object.
Physics 201: Lecture 16, Pg 24
Work and Energy

An object moves along a line under the influence of a single
force. The area under the force vs. position graph represents
(a) the impulse delivered to the object
(b) the work done on the object.
(c) the change in the velocity of the object.
(d) the momentum of the object.
Physics 201: Lecture 16, Pg 25
Concept problem

At high speeds the drag force of the air is found to be
proportional to the square of a car’s speed. Assume that at 60
mph that 100% of car’s power is being used against wind
resistance (i.e., there are no other non-conservative forces.) In
terms of the ratio P(120 mph) / P (60 mph), how much more
power will the car’s engine need to provide if this car is to travel
at 120 mph?
A.
2
23/2
4
25/2
8
B.
C.
D.
E.
P = F v = v3 so ratio is 8
Physics 201: Lecture 16, Pg 26
Work and Energy


A block slides along a frictionless surface before colliding with a spring.
The block is brought momentarily to rest by the spring after traveling
some distance. The four scenarios shown in the diagrams below are
labeled with the mass of the block, the initial speed of the block, and
the spring constant.
Rank the scenarios in order of the distance the block travels, listing the
largest distance first.
(a) B , A , C = D
(b) B , C , A , D
(c) B , C = D , A
(d) C = B, A , D
(e) C = B = D , A
Physics 201: Lecture 16, Pg 27
Work and Energy


A block slides along a frictionless surface before colliding with a spring.
The block is brought momentarily to rest by the spring after traveling
some distance. The four scenarios shown in the diagrams below are
labeled with the mass of the block, the initial speed of the block, and
the spring constant.
Rank the scenarios in order of the distance the block travels, listing the
largest distance first.
(a) B , A , C = D
(b) B , C , A , D
(c) B , C = D , A
(d) C = B, A , D
(e) C = B = D , A
Physics 201: Lecture 16, Pg 28
Work and Forces

A 25.0 kg chair is pushed 2.00 m at constant speed along
a horizontal surface with a constant force acting at 30.0
degrees below the horizontal. If the friction force between
the chair and the surface is 55.4 N, what is the work done
by the pushing force?
(a) 85 J
(b) 98 J
(c) 111 J
(d) 113 J
(e) 128 J
Physics 201: Lecture 16, Pg 29
Work and Forces

A 25.0 kg chair is pushed 2.00 m at constant speed along
a horizontal surface with a constant force acting at 30.0
degrees below the horizontal. If the friction force between
the chair and the surface is 55.4 N, what is the work done
by the pushing force?
(a) 85 J
(b) 98 J
(c) 111 J
(d) 113 J
(e) 128 J
Physics 201: Lecture 16, Pg 30
Work and Power

A 100 kg elevator is carrying 6 people, each weighing 70
kg. They all want to travel to the top floor, 75 m from the
floor they entered at. How much power will the elevator
motor supply to lift this in 45 seconds at constant speed?
(a) 1.2 · 102 W
(b) 7.0 · 102 W
(c) 8.7 · 102 W
(d) 6.9 · 103 W
(e) 8.5 · 103 W
Physics 201: Lecture 16, Pg 31
Work and Power

A 100 kg elevator is carrying 6 people, each weighing 70
kg. They all want to travel to the top floor, 75 m from the
floor they entered at. How much power will the elevator
motor supply to lift this in 45 seconds at constant speed?
(a) 1.2 · 102 W
(b) 7.0 · 102 W
(c) 8.7 · 102 W
(d) 6.9 · 103 W
(e) 8.5 · 103 W
Physics 201: Lecture 16, Pg 32
Newton’s Laws

Two sleds are hooked together in tandem. The front sled
is twice as massive as the rear sled. The sleds are pulled
along a frictionless surface by a force F, applied to the
more massive sled. The tension in the rope between the
sleds is T. Determine the ratio of the magnitudes of the
two forces, T/F.
(a) 0.33
(b) 0.50
(c) 0.67
(d) 1.5
(e) 2.0
(f) 3.0
Physics 201: Lecture 16, Pg 33
Newton’s Laws

A factory worker raises a 100. kg crate at a constant rate
using a frictionless pulley system, as shown in the
diagram. The mass of the pulleys and rope are negligible.
(assume g= 9.8 m/s2)
With what force is the worker pulling down on the rope?
(a) 245 N
(b) 327 N
(c) 490 N
(d) 980 N
(e) 1960 N
Physics 201: Lecture 16, Pg 34
Work and Energy

A 6.0 kg block is pushed up against an ideal Hooke’s law spring
(of spring constant 3750 N/m ) until the spring is compressed a
distance x. When it is released, the block travels along a track
from one level to a higher level, by moving through an
intermediate valley (as shown in the diagram). The track is
frictionless until the block reaches the higher level. There is a
frictional force stops the block in a distance of 1.2 m. If the
coefficient of friction between the block and the surface is 0.60,
what is x ? (Let g = 9.81 m/s2 )
(a) 0.11 m
(b) 0.24 m
(c) 0.39 m
(d) 0.48 m
(e) 0.56 m
Physics 201: Lecture 16, Pg 35
Work and Energy

A 6.0 kg block is pushed up against an ideal Hooke’s law spring
(of spring constant 3750 N/m ) until the spring is compressed a
distance x. When it is released, the block travels along a track
from one level to a higher level, by moving through an
intermediate valley (as shown in the diagram). The track is
frictionless until the block reaches the higher level. There is a
frictional force stops the block in a distance of 1.2 m. If the
coefficient of friction between the block and the surface is 0.60,
what is x ? (Let g = 9.81 m/s2 )
(a) 0.11 m
(b) 0.24 m
(c) 0.39 m
(d) 0.48 m
(e) 0.56 m
Physics 201: Lecture 16, Pg 36
Conceptual Problem
A bird sits in a birdfeeder suspended from a
tree by a wire, as shown in the diagram at
left.
Let WB and WF be the weight of
the bird and the feeder
respectively. Let T be the
tension in the wire and N be the
normal force of the feeder on
the bird. Which of the following
free-body diagrams best
represents the birdfeeder?
(The force vectors are not
drawn to scale and are only
meant to show the direction, not
the magnitude, of each force.)
Physics 201: Lecture 16, Pg 37
Conceptual Problem
A block is pushed up a 20º ramp by a 15 N force which may
be applied either horizontally (P1) or parallel to the ramp (P2).
How does the magnitude of the normal force N depend on the
direction of P?
(A) N will be smaller if P is
horizontal than if it is parallel
the ramp.
(B) N will be larger if P is
horizontal than if it is parallel
to the ramp.
(C) N will be the same in both
cases.
(D) The answer will depend on
the coefficient of friction.
20°
Physics 201: Lecture 16, Pg 38
Conceptual Problem
A cart on a roller-coaster rolls down the track shown below.
As the cart rolls beyond the point shown, what happens to
its speed and acceleration in the direction of motion?
A. Both decrease.
B. The speed decreases, but
the acceleration increases.
C. Both remain constant.
D. The speed increases, but
acceleration decreases.
E. Both increase.
F. Other
Physics 201: Lecture 16, Pg 39
Sample Problem
A 200 kg wood crate sits in the back of a truck. The coefficients
of friction between the crate and the truck are μs = 0.9 and μk =
0.5.
The truck starts moving up a 20° slope. What is the maximum
acceleration the truck can have without the crate slipping out
the back?
 Solving:
 Visualize the problem, Draw a picture if necessary
 Identify the system and make a Free Body Diagram
 Choose an appropriate coordinate system
 Apply Newton’s Laws with conditional constraints (friction)
 Solve

Physics 201: Lecture 16, Pg 40
Sample Problem


You have been hired to measure the coefficients of friction
for the newly discovered substance jelloium. Today you will
measure the coefficient of kinetic friction for jelloium sliding
on steel. To do so, you pull a 200 g chunk of jelloium across
a horizontal steel table with a constant string tension of 1.00
N. A motion detector records the motion and displays the
graph shown.
What is the value of μk for jelloium on steel?
Physics 201: Lecture 16, Pg 41
Sample Problem
S Fx =ma = F - ff = F - mk N = F - mk mg
S Fy = 0 = N – mg
mk = (F - ma) / mg & x = ½ a t2  0.80 m = ½ a 4 s2
a = 0.40 m/s2
mk = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46
Physics 201: Lecture 16, Pg 42
Exercise: Newton’s 2nd Law
A force of 2 Newtons acts on a cart that is initially at rest
on an air track with no air and pushed for 1 second.
Because there is friction (no air), the cart stops
immediately after I finish pushing.
It has traveled a distance, D.
Force
Cart
Air Track
Next, the force of 2 Newtons acts again but is
applied for 2 seconds.
A.
B.
The new distance the cart moves relative to D
is:
C.
D.
8 x as far
4 x as far
2 x as far
1/4 x as far
Physics 201: Lecture 16, Pg 43
Exercise: Solution
Force
Cart
Air Track
We know that under constant acceleration,
Dx = a (Dt)2 /2
(when v0=0)
Here Dt2=2Dt1,
F2 = F1  a2 = a1
1
aDt 22 
Dx2 2
2Dt1 2


4
2
Dx1 1 aDt 2
Dt1
1
2
(B) 4 x as long
Physics 201: Lecture 16, Pg 44
Sample exam problem
A small block moves along a frictionless
incline which is 45° from horizontal.
Gravity acts down at 10 m/s2. There
is a massless cord pulling on the
block. The cord runs parallel to the
incline over a pulley and then straight
down. There is tension, T1, in the cord
which accelerates the block at 2.0
m/s2 up the incline. The pulley is
suspended with a second cord with
tension, T2.
A. What is the tension magnitude, T1, in the 1st cord?
B. What is the tension magnitude,T2, in the 2nd cord?
Physics 201: Lecture 16, Pg 45
Sample exam problem
a = 2.0 m/s2 up the incline.
What is the tension magnitude, T1, in
the 1st cord?
Use a FBD!
Along the block surface
S Fx = m ax = -mg sin q + T
T = 5 x 2 N + 5 x 10 x 0.7071 N
= (10 + 35) N = 45 N
Physics 201: Lecture 16, Pg 46
Sample exam problem
a = 0.0 m/s2 at the pulley.
What is the tension magnitude,T2, in
the 2nd cord?
Use a FBD!
Physics 201: Lecture 16, Pg 47
Sample exam problem

You have a 2.0 kg block that moves on a linear path on a horizontal
surface. The coefficient of kinetic friction between the block and the
path is μk. Attached to the block is a horizontally mounted massless
string as shown in the figure below. The block includes an
accelerometer which records acceleration vs. time. As you increase
the tension in the rope the block experiences an increasingly positive
acceleration. At some point in time the rope snaps and then the block
slides to a stop (at a time of 10 seconds). Gravity, with g = 10 m/s2,
acts downward.
Physics 201: Lecture 16, Pg 48
Sample exam problem
A. At what time does the string break and, in one sentence,
explain your reasoning?
B. What speed did the block have when the string broke?
C. What is the value of μk?
D. Using μk above (or a value of 0.25 if you don’t have one),
what was the tension in the string at t = 2 seconds?
Physics 201: Lecture 16, Pg 49
Sample exam problem
B. What speed did the block have when the string broke?
Don’t know initial v (t=0) so can’t integrate area at t < 4 sec.
vf = 0 m/s and from t = 4 to 10 sec (6 second) a = - 2 m/s2
0 = vi + a t = vi – 2 x 6 m/s  vi = 12 m/s
Physics 201: Lecture 16, Pg 50
Sample exam problem
C. What is the value of μk?
Use a FBD!
S Fx = m ax = - fk = - μk N
S Fy = 0 = mg – N  N = mg
So m ax = - fk = - μk mg  μk = - ax / g = - (-2)/10 = 0.20
Physics 201: Lecture 16, Pg 51
Sample exam problem
D. What was the tension in the string at t = 2 seconds?
Again a FBD!
S Fx = m ax = - fk + T
S Fy = 0 = mg – N 
N = mg
T = fk + m ax = (0.20 x 2 x 10 + 2 x 3 ) N = 10 N
Physics 201: Lecture 16, Pg 52
Sample exam problem
A 5.0 kg block with an equilateral triangular
cross-section lies as shown in a 60°
frictionless groove. Gravity, with g = 10.
m/s2, acts downward.
You may use the following relationships:
cos 60° = sin 30° =0.50
cos 30° = sin 60° =0.87.
A. Draw a free body diagram showing all the
forces acting on the block.
B. What are the magnitudes of the normal
forces on the block associated with the
two sides of contact (A to B and B to C)?
N
N’
q q
mg
Use a FBD! ( q = 30°)
S Fx = 0 = -N cos 30° + N’ cos 30°  N = N’
S Fy = 0 = -mg + N sin 30° + N’ sin 30°  N = 50 N
Physics 201: Lecture 16, Pg 53
This is why wedges are good for splitting things