Download Physics 101: Lecture 04

Physics 101: Lecture 04
Exam
I
Kinematics + Dynamics
 Today’s
lecture will cover Textbook
Chapter 4
Physics 101: Lecture 4, Pg 1
What’s Most Difficult
I’m a little slow dusting off the trig skills.
Identifying all forces involved with free body diagrams.
…I think I just need some more practice walking through
the thought process required to solve the problems.
I am having a hard time with vectors in this class…
The most difficult problems right now are 2 dimension analysis problems
where you don’t know 2 variables so you have to end up substituting in one
variable for the other and manipulating equations.
Everything :(
I LOVE Free Body Diagrams.
The lectures don’t seem to explain much, just random letters and numbers…
Physics 101: Lecture 4, Pg 2
05
Review
 Kinematics
: Description of Motion
Position
Displacement
Velocity v = Dx / Dt
» average
» instantaneous
Acceleration a = Dv / Dt
» average
» Instantaneous
Physics 101: Lecture 4, Pg 3 07
Preflight 4.1
…most trouble with deciphering graphs…
(A)
(B)
(C)
•Which x vs t plot shows positive acceleration?
90% got this correct!!!!
“The slope of the graph is getting increasingly positive; the velocity is increasing
over time.”
Physics 101: Lecture 4, Pg 4 08
Equations for Constant
Acceleration
(text, page 108)
x (meters)
200
150
100
50
0



x = x0 + v0t + 1/2
at2

v = v0 + at

v2 = v02 + 2a(x-x0)

Dx = v0t + 1/2
0
5
10
t (seconds)
15
20
5
10
t (seconds)
15
20
5
10
t (seconds)
15
20
v (m/s)
at2
Dv = at
20
15
10
v2 = v02 + 2a Dx
Lets derive this…
5
0
0
2
a (m/s )
2
1.5
1
0.5
0
0
Physics 101: Lecture 4, Pg 5
10
Equations for Constant
Acceleration
(text, page 108)
x (meters)
200
150
100
50
0



x = x0 + v0t + 1/2
at2

v = v0 + at

v2 = v02 + 2a(x-x0)

Dx = v0t + 1/2
0
5
10
t (seconds)
15
20
5
10
t (seconds)
15
20
5
10
t (seconds)
15
20
v (m/s)
at2
Dv = at
20
15
10
v2 = v02 + 2a Dx
5
0
0
2
a (m/s )
2
1.5
1
0.5
0
0
Physics 101: Lecture 4, Pg 6
14
Kinematics Example

A car is traveling 30 m/s and applies its breaks to stop
after a distance of 150 m. How fast is the car going after
it has traveled ½ the distance (75 meters) ?
A) v < 15 m/s
B) v = 15 m/s
C) v > 15 m/s
Physics 101: Lecture 4, Pg 9
18
Acceleration ACT
An car accelerates uniformly from rest. If it travels a distance
D in time t then how far will it travel in a time 2t ?
A. D/4
B. D/2
C. D
Demo…
D. 2D
Correct x=1/2 at2
E. 4D
Follow up question: If the car has speed v at time t then
what is the speed at time 2t ?
A. v/4
B. v/2
C. v
D. 2v
Correct v=at
E. 4v
Physics 101: Lecture 4, Pg 10 24
Newton’s Second Law SF=ma
Physics 101: Lecture 4, Pg 12
ACT

A force F acting on a mass m1 results in an acceleration
a1.The same force acting on a different mass m2 results
in an acceleration a2 = 2a1. What is the mass m2?
m1
F
(A)
2m1
a1
m2
F
(B) m1
(C)
a2 = 2a1
1/2 m1
• F=ma
• F= m1a1 = m2a2 = m2(2a1)
• Therefore, m2 = m1/2
• Or in words…twice the acceleration means half the mass
Physics 101: Lecture 4, Pg 13
29
Example:
A tractor T (m=300Kg) is pulling a trailer M (m=400Kg). It
starts from rest and pulls with constant force such that there is
a positive acceleration of 1.5 m/s2. Calculate the horizontal
force on the tractor due to the ground.
X direction: Tractor
SF = ma
Fw – T = mtractora
Fw = T+ mtractora
X direction: Trailer
SF = ma
T = mtrailera
y
x
FW = 1050 N
N
N
Fw
T
T
W
W
Combine:
Fw = mtrailera + mtractora
Fw = (mtrailer+mtractor ) a
Physics 101: Lecture 4, Pg 14
35
Net Force ACT
Compare Ftractor the net force on the tractor,
with Ftrailer the net force on the trailer from
the previous problem.
SF = m a
A) Ftractor > Ftrailor
Ftractor = mtractor a
B) Ftractor = Ftrailor
2)
=
(300
kg)
(1.5
m/s
C) Ftractor < Ftrailor
= 450 N
Ftrailer = mtrailer a
= (400 kg) (1.5 m/s2)
= 600 N
Physics 101: Lecture 4, Pg 15
38
Pulley Example

Two boxes are connected by a string over a frictionless pulley. Box 1
has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8
meters above the table, how long does it take to hit the table.
y
•Compare the acceleration of boxes 1 and 2
A) |a1| > |a2|
B) |a1| = |a2|
x
C) |a1| < |a2|
1) T - m1 g = m1 a1
2) T – m2 g = -m2 a1
1
T
2) T = m2 g -m2 a1
1) m2 g -m2 a1 - m1 g = m1 a1
a1 = (m2 – m1)g / (m1+m2)
1
T
2
2
m
m24,gPg 16
1g 101: Lecture
Physics
42
Pulley Example

Two boxes are connected by a string over a frictionless pulley. Box 1
has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8
meters above the table, how long does it take to hit the table.
y
•Compare the acceleration of boxes 1 and 2
A) |a1| > |a2|
B) |a1| = |a2|
a1 = (m2 – m1)g / (m1+m2)
a = 2.45 m/s2
Dx = v0t + ½ a t2
Dx = ½ a t2
t = sqrt(2 Dx/a)
t = 0.81 seconds
x
C) |a1| < |a2|
1
T
1
T
2
2
m
m24,gPg 17
1g 101: Lecture
Physics
46
Summary of Concepts
• Constant Acceleration

x = x0 + v0t + 1/2 at2

v = v0 + at

v2 = v02 + 2a(x-x0)
I feel that if there were
suggested homework
problems from the book,
I would have a lot more
practice with the
concepts we’re learning.
•F=ma
– Draw Free Body Diagram
– Write down equations
– Solve
•Ch 2 Problems: 1,5,7,11,13,17,29,49,69
•Ch 3 Problems: 5,13,33,47,57,65,67
Physics 101: Lecture 4, Pg 18