Download 25 newtons antics 35s 15s - lindsey

Newton’s Antics
Feb. 20th, 2009
Ridiculous Reminders
• Take Motion Test – Progress Reports Due
Monday!
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Melissa R.
Jordan S.
Kim S.
Mackenzie T.
Gau Yee
Newton’s Laws of Motion Wkst #2
• Please take out the 35’s
•
Please follow along, answer questions, ask questions, basically be a student – LEARN!
Problem #1
If a massive truck and a smaller car going the same speed have a headon collision, which vehicle will experience the greater force? Which
vehicle will experience greater acceleration?
forces must be equal because of
action/reaction
Car has a greater acceleration
F=F
M
a
=M
a
Problem #2
What is terminal velocity? Does a heavy or light person have a higher
terminal velocity? Who reaches the ground first if they parachute at
the same time?
force of friction = force of gravity
air resistance = weight
– acceleration = 0 m/s2
– Fnet = 0 N
Terminal velocity is when
• Heavy person has a higher terminal velocity
• Heavy person
Problem #3
• A force is needed to accelerate a brick. If twice the force
is used to push the same mass, how does acceleration
change?
• F = ma
• 2F = m(?a)
• = 2a
Problem #4
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A force is needed to accelerate a brick. If 3 bricks (3
times the mass) are pushed with the same force, how
does acceleration change?
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F=ma
F=3m(?a)
= 1/3 a
Problem #5
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A man drives his snowmobile across Lake Wausau in the late spring when
the ice is getting thin. The man and his snowmobile have a combined
mass of 300 kg. The area of contact is 1.5 meters by 0.5 meters. The ice
can only withstand a pressure of 3000 N/m2. Does the man fall through the
ice?
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Area = Length x width
Area = 1.5 m x .5 m = 0.75 m2
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P = Force / Area
3000 N/m2 = F / 0.75 m2
F = 2250 N
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Force of Snowmobile and Person
F = 300 kg (9.8 m/s2) = 2940 N
Snowmobiler
Falls through
the Ice
Problem #6
• A) Which exerts a greater force on the table, a book lying flat or the
book standing on its end? b) Which applies a greater pressure? c)
Calculate the pressure of each, if the book has a mass of 1.6 kg and
measures 0.26 m x 0.21 m x 0.04 m.
A) same
B) P = F/A
C) Flat Book
P = F/A
P = 1.6kg (9.8 m/s2) /
(.26 m x .21 m)
P = 287.2 N/m2
on edge
C) On End Book
P = F/A
P = 1.6kg (9.8 m/s2) /
(.04 m x .21 m)
P = 1866.7 N/m2
Problem #7
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A 5 N force upward and a 5 N force to the right act on an object.
What is the net force? A 3 N, 7 N, and 10 N force act on an object
as shown. What is the net force?
a2
52
+
b2
+
52
=
c2
a2 + b2 = c2
=
c2
72 + 72 = c2
25 + 25 =
49 + 49 = c2
c2
c = 9.89 N
c = 7.07 N
5
7
5
7
Problem #8
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What creates the ability for planes to fly?
Explain and draw a diagram.
Problem #9
• A roller coaster with 20 passengers descends a hill at an angle of
45°. If the rollercoaster itself has a mass of 600 kg and you assume
each passenger weighs about 51 kg. What is the force pulling them
down the hill?
Fg = [(20 * 51 kg)+600 kg] (9.8 m/s2)
Fg = 15,876 N
FN = Fg * cos 45˚ = 11,226 N
Fp = Fg * sin 45˚ = 11,226 N
Problem #10
•
At Sea World a 900 kg polar bear slides down a wet slide at an
angle of 25.0°. The coefficient of friction between the bear and
the slide is 0.050. What is the force of friction that opposes the
bear’s motion?
FN = 900 kg * 9.8 m/s2* cos 25˚
FN = 7994 N
Ff = 0.050 * 7994 N
Ff = 400 N
Friction Lab 15s
1. In which direction does the frictional force act as you are
walking forward?
friction is defined as the force that opposes motion therefore
as you walk forward, friction acts backward
2. Can the coefficient of friction be greater than 1.0? Why?
In most relevant situations the FN>Ff therefore μ<1
3. If the mass of your shoe increased, what happens to the
coefficient of friction?
μ would be the same
4. When a wood block is pulled across the table at a constant
velocity, how does the force of you pulling compare to the
force of friction? Is the block accelerating? What is the net
force on the block?
Fp=Ff
constant velocity means no acceleration
if there is no acceleration then the net force must be zero
Practice Problem
• Billy rollerblades across the floor. He weighs 560
N and the force of friction is 112 N. What is the
coefficient of friction?
Fn = 560 N
Ff = 112 N
Ff = μ Fn
112 N = μ (560 N)
μ = 0.20
Practice Problem
• A student on a cart is pushed with a force of 10
N. The student and cart accelerate from rest to 2
m/s in a time of 10 seconds. What is the mass of
the student and cart?
F = 10 N
a=?
Δv = 2 m/s – 0 m/s = 2 m/s
t = 10 s
a = 2 m/s / 10 s = 0.20 m/s2
F= m*a
10 N = m * 0.20 m/s2
m = 50 kg
Practice Problem
• Two people are pulling a boat against the current of a
river. Each person exerts a 600 N force at a 30˚ angle
relative to the boat. If the boat moves with a constant
velocity, find the force of the current.
Force of Person 1
F1 = Fa cos θ
F1 = 600 N* cos 30˚
Force of Person 2
F1 = F2
F2 = 520 N
FT = 1040 N
F1 = 520 N
Fcurrent = 1040 N
Practice Problem
• A hockey puck is given an initial speed of 20.0 m/s
on a frozen pond. The puck remains on the ice and
slides 120 m before coming to rest in 12 seconds.
Determine the coefficient of friction between the
puck and ice.
Finitial = Ffinal
m*ai = μ Fn
a = Δv/Δt
m*ai = μ*m*g
a = (-20 m/s)/ 12 s
ai = μ*g (solve on your own)
a = -1.67 m/s2
μ = 0.170
Practice Problem
• The combined weight of a crate and dolly is
300 N. If a person pulls on a rope with a
constant force of 20.0 N, what is the
acceleration of the system if no frictional
forces apply?
w = m*g
F = ma
300 N = m *9.8 m/s2
20 N = 30.6 kg * a
m = 30.6 kg
a = 0.654 m/s2
Reminders
• PotW due Monday
• 36’s Due Tuesday
• Test on Thursday
• Packet- make sure 13s, 14s, 15s, 33s,
34s, 35s, 36s are complete to be turned in
on Thursday