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Physics Review
By: Kellie Krueger
Velocity, Acceleration, and
Displacement
Velocity
V
l i – Rate
R
off change
h
off position.
ii
Change in position / change in time
∆x/∆t
vf² = vi² + 2a∆x
vf = at + vi
„
Acceleration – increase or decrease in velocity
velocity.
Change in velocity / change in time
∆v/∆t
„
Displacement – the vector quantity that defines the distance and
direction between 2 positions.
∆x = ½ at2 + vit
„
Example Problem
A bike travels at a constant speed of 4.0
m/s for 5 s. How far does it go?
1.
v = ∆x/∆t
2.
4 0m/s = ∆x / 5s
4.0m/s
3.
4.0(5s)
4.
∆x = 20m
Example Problem 2
A car can accelerate from rest to speed of
28m/s in 20s.
a) what is the average acceleration of the
car?
a = ∆v/∆t
a = 28m/s
/ / 20s
a = 1.4m/s
/ 2
Example Problem 3
A 65 kg person dives into the water from a 10m
platform.
„ What is her speed as she enters the water?
Given:
„ Mass = 65 kg therefore Fg = -650N
„ ∆x = 10m
Trying to find:
„ Speed
Force Diagram
Normal Force
65 kg
Force of gravity
= -650N
Solving the Problem
vf² = vi² + 2a∆x
vf² = (0m/s)² + 2(9.8m/s²)(10m)
vf = 14m/s
Force, Mass, and Acceleration
The sum of the forces is equal to mass
multiplied by acceleration.
F = ma
Example Problem
An elevator is moving up at a constant velocity of
2.5m/s. The mass of the man = 85 kg.
„ What force does the floor exert on the man?
Given:
G
e
„ Because of constant velocity (2.5m/s), a =
0m/s²
„ Mass
M
= 85kg,
85k therefore,
th f
Fg = -850N
850N
Trying to find:
„ Normal Force
Force Diagram
Normal Force
Gravity Force = -850N
Solving the Problem
F = ma
-850N = (85kg)(0m/s²)
Normall Force = 850N
Newton’s 2nd Law and Friction
Coefficient of friction: µ
FF= µ ‫ ٭‬FN
Example Problem
A sled weighing 300N is moved at a
constant speed over a horizontal floor
by a force of 50N applied parallel to the
floor.
„ What is the coefficient of friction
between the sled and the floor?
Force Diagram
Normal
Force: 300N
Friction
Force: ?
Applied
Force: 50N
Gravity Force: -300N
Givens and Unknowns
Given:
„ Since there is a constant speed,
p
,a=0
„ Applied force = 50N
„ Gravity force = -300N
300N
„ Normal force = 300N
Trying to find:
„ Coefficient of friction
Solving the Problem
µ = FF/FN
µ = 50N/300N
µ = .167
Problem 2
What would be the acceleration of the
sled if µ = 0?
F = ma
50N = 30kg(a)
a = 1.67 m/s²
„