101: Physics Lecture 04
Exam I
Kinematics + Dynamics
Todays lecture will cover Textbook Chapter 4
Physics 101: Lecture 4, Pg 1
Review
Kinematics : Description of Motion position displacement Velocity v = x / t
average instantaneous
Acceleration a = v / t
average Instantaneous
Physics 101: Lecture 4, Pg 2
Equations for Constant Acceleration (text, page 80)
x = x0 + v0t + 1/2 at2 v = v0 + at v2 = v02 + 2a(x-x0) x = v0t + 1/2 at2 v = at v2 = v02 + 2a x
x (meters) 200
150
100
50
0
0
5
10 t (seconds)
15
20
v (m/s) 20
15
10
5
0
0
5
Lets derive this
10 t (seconds)
15
20
a (m/s ) 2
2
1.5
1
0.5
0
0
5
Physics 101: Lecture 4, Pg 3
10 t (seconds)
15
20
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 the distance (75 meters) ? A) v < 15 m/s B) v = 15 m/s C) v > 15 m/s
Physics 101: Lecture 4, Pg 4
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 D. 2D 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 E. 4v
Physics 101: Lecture 4, Pg 5
ACT
A force F acting on a mass m1 results in an a1.The acceleration same force acting on a different mass m2 results in an acceleration a2 = 2a1. What is the mass m2?
F (1) m1 a1 (2) m1 F (3) m2
a2 = 2a1
2m1
1/2 m1
Physics 101: Lecture 4, Pg 6
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.
y x
Physics 101: Lecture 4, Pg 7
Physics 101: Lecture 4, Pg 8
Net Force ACT
Compare Ftractor the net force on the tractor, with Ftrailer the net force on the trailer from the previous problem. A) Ftractor > Ftrailor B) Ftractor = Ftrailor C) Ftractor < Ftrailor
Physics 101: Lecture 4, Pg 9
Pulley Example
Two boxes are connected by a stri...