Newton’s Antics

Feb. 20th, 2009

Ridiculous Reminders

• Take Motion Test – Progress Reports Due Monday!

• 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 head-on collision, which vehicle will experience the greater force? Which vehicle will experience greater acceleration?

F = F

M a = M a

forces must be equal because of action/reaction

Car has a greater acceleration

Problem #2What 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?

Terminal velocity is when force of friction = force of gravity air resistance = weight

– acceleration = 0 m/s2

– Fnet = 0 N

• 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

• 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?

• F=ma

• F=3m(?a)

• = 1/3 a

Problem #5• 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?

• Area = Length x width• Area = 1.5 m x .5 m = 0.75 m2

• P = Force / Area• 3000 N/m2 = F / 0.75 m2

• F = 2250 N

• 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 on edge

C) Flat BookP = F/A

P = 1.6kg (9.8 m/s2) / (.26 m x .21 m)P = 287.2 N/m2

C) On End BookP = F/A

P = 1.6kg (9.8 m/s2) / (.04 m x .21 m)

P = 1866.7 N/m2

Problem #7

• 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?

5

5

7

7

a2 + b2 = c2

52 + 52 = c2

25 + 25 = c2

c = 7.07 N

a2 + b2 = c2

72 + 72 = c2

49 + 49 = c2

c = 9.89 N

Problem #8

• 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 15s1. 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˚

F1 = 520 N

Force of Person 2F1 = F2

F2 = 520 NFT = 1040 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

a = Δv/Δt

a = (-20 m/s)/ 12 s

a = -1.67 m/s2

m*ai = μ Fn

m*ai = μ*m*g

ai = μ*g (solve on your own)

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

300 N = m *9.8 m/s2

m = 30.6 kg

F = ma

20 N = 30.6 kg * a

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