Today
Kinematics + Dynamics
Physics 101: Lecture 5, Pg 1
Example: 2 block & pulley
Two blocks are connected by a rope that passes over a pulley.
The 4kg block is held at rest on a frictionless surface. After being
released what is the acceleration of the blocks?
4kg
2kg
Physics 101: Lecture 5, Pg 2
Another method
4kg
2kg
2kg
4kg
Physics 101: Lecture 5, Pg 3
Equations for Constant
Acceleration
x (meters)
200
150
100
50
0
0
5
10
t (seconds)
15
20
5
10
t (seconds)
15
20
5
10
t (seconds)
15
20
v (m/s)
x
v
= x0 + v0t + 1/2
at2
= v0 + at
20
15
10
5
 v2
= v02 + 2a(x-x0)
0
0
2
a (m/s )
2

x = v0t + 1/2 at2

v = at

v2 = v02 + 2a x
1.5
1
0.5
0
0
Physics 101: Lecture 5, Pg 4
10
Example: Braking car

A car is traveling 30 m/s and applies its brakes to stop
with a constant acceleration in 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 5, Pg 5
18
Example: ball thrown up
•Find time to go up:
•Find max height
Vo=25m/s
Physics 101: Lecture 5, Pg 6
Example: ball thrown up
•Find time to come down:
•Find velocity right before it hits ground
Vo=25m/s
Physics 101: Lecture 5, Pg 7
Example: ball thrown from cliff
•Find velocity right before it hits ground
Vo=25m/s
60m
•Find total time in air
Physics 101: Lecture 5, Pg 8
Example: ball thrown from cliff
•Find velocity 30m above ground
Vo=25m/s
What is the vertical
displacement
A. 30 m
B. 61.9 m
C. -30 m
D. -61.9 m
E. 93.8 m
60m
Physics 101: Lecture 5, Pg 9
Summary of Concepts
• Constant Acceleration

x = x0 + v0t + 1/2 at2

v = v0 + at

v2 = v02 + 2a(x-x0)
•F=ma
– Draw Free Body Diagram
– Write down equations
– Solve
Physics 101: Lecture 5, Pg 10