Transcript Ch6

Ch6.1 - Forces
Newton's 1st Law - object in motion will remain in that motion,
an object at rest stays at rest.
- Will not change unless a net force acts on it.
- Rest is just a special case of motion.
Newton's 1st Law - object in motion will remain in that motion,
an object at rest stays at rest.
- Will not change unless a net force acts on it.
- Rest is just a special case of motion.
Inertia - tendency of an object to resist change.
Newton's 1st Law - object in motion will remain in that motion,
an object at rest stays at rest.
- Will not change unless a net force acts on it.
- Rest is just a special case of motion.
Inertia - tendency of an object to resist change.
Force - push or pull
Contact force - touches object
Field force - long range, acts at a distance
Equilibrium - net force on an object is zero
(object can be at rest or moving at constant speed)
Free body diagram - drawing of object with all forces labeled on it.
Net Force - what’s left over after all forces we added up.
Force is a vector, so the vectors have to be added vectorally!
Ex1) What is the net force in each:
a. F1
F2
b. F1
F2
c. F1
F2
Newton’s 2nd Law:
F = (m)(a) 
Units of force:
a
Newton’s 2nd Law:
F = (m)(a) 
Units of force: kg 
Fnet
a
m
m
= Newtons
2
s
Types of Forces:
Fg = Force of Gravity
(weight)
FN = Normal force
Don’t confuse Fnet with FN
FT = Tension
Ex 2) A rope lifts a bucket, its speed is increasing. Draw and label forces.
What is net force?
Air Drag
- Friction with the air DOES
depend on speed
(Faster you fall, the more air
has to get out of your way.)
- DOES depend on surface area.
(Same reason)
1.
2.
3.
(BASE jumper)
HW#1) Draw free body diagrams for:
a. A book held stationary in hand.
b. Book pushed across a table at constant speed.
HW #4) A 225N force is exerted on a crate to the North.
A 165N force is also exerted on the crate, but east.
Find the net force, both magnitude and direction.
CH 6 HW #1: 1-8
Lab6.1 – Inertia
- due tomorrow
- Ch6 HW#1 due at beginning of period
Ch6 HW#1 1- 8
1. Draw:
c. A book on the desk while you hand is
pushing down on it.
d. A ball, just as the string holding it breaks.
2. 2 horiz forces 225N and 165N pushing on crate in same direction.
Find Fnet.
FN
F = 165N
F = 225N
Fg
Ch6 HW#1 1- 8
1. Draw:
c. A book on the desk while you hand is
pushing down on it.
d. A ball, just as the string holding it breaks.
FN
Fg
FHand
c. Fnet = Fg + FHand – FN
d. Fnet = Fg
Fg
2. 2 horiz forces 225N and 165N pushing on crate in same direction.
Find Fnet.
FN
F = 165N
F = 225N
Fg
Ch6 HW#1 1- 8
1. Draw:
c. A book on the desk while you hand is
pushing down on it.
d. A ball, just as the string holding it breaks.
c. Fnet = Fg + FHand – FN
FN
Fg
FHand
d. Fnet = Fg
Fg
2. 2 horiz forces 225N and 165N pushing on crate in same direction.
Find Fnet.
FN
F = 165N
F = 225N
Fnet = 165N + 225N
= 390N
Fg
3. 2 horiz forces 225N and 165N pushing on crate in opp directions.
Find Fnet.
FN
F = 165N
F = 225N
Fg
4. A 225N force is exerted on a crate to the North.
A 165N force is also exerted on the crate, but east.
Find the net force, both magnitude and direction.
3. 2 horiz forces 225N and 165N pushing on crate in opp directions.
Find Fnet.
FN
F = 165N
F = 225N
Fnet = 225N – 165N
= 60N to the right
Fg
4. A 225N force is exerted on a crate to the North.
A 165N force is also exerted on the crate, but east.
Find the net force, both magnitude and direction.
3. 2 horiz forces 225N and 165N pushing on crate in opp directions.
Find Fnet.
FN
F = 165N
F = 225N
Fnet = 225N – 165N
= 60N to the right
Fg
4. A 225N force is exerted on a crate to the North.
A 165N force is also exerted on the crate, but East.
Find the net force, both magnitude and direction.
Fnet =
225 2  165 2
= 279N
1 
225 

  tan 
  54
 165 
5. Calc the force you exert on the floor while you stand. (1 lb = .454kg)
Does the force change if you lie down?
150 lb
6. A cable pulls a crate at constant speed across the floor.
HW #7) A skydiver falls downward thru the air, accelerating
8. A rope lowers a bucket at constant speed.
HW #7) A skydiver falls downward thru the air, accelerating
Fa.r.
Fg
8. A rope lowers a bucket at constant speed.
FT
Fnet = Fg – FT
0 = Fg – FT
Fg = FT
Fg
Ch6.2 – Mass vs. Weight
- All objects in free fall accelerate at g = 9.8 m/s2
since
F = m.a
Ch6.2 – Mass vs. Weight
- All objects in free fall accelerate at g = 9.8 m/s2
since
F = m.a
Force of gravity: Fg = m.g
Weight – measure of the force of gravity (Newtons, N)
Mass – amount of matter present (kilograms, kg.)
Ex1) My mass is 80 kg,
a. What’s my weight?
b. What’s my mass on the moon?
c. If gravity on the moon is 1.67m/s2, what’s my weight on the moon?
If you are accelerating up or down (like in an elevator),
your weight changes.
- If it accelerates upward, you squish the scale more
& you appear to weigh more.
F = m(g + a)
If you are accelerating up or down (like in an elevator),
your weight changes.
- If it accelerates upward, you squish the scale more
& you appear to weigh more.
F = m(g + a)
- If it accelerates downward, the scale expands
& you apparently weigh less.
F = m(g – a)
If you are accelerating up or down (like in an elevator),
your weight changes.
- If it accelerates upward, you squish the scale more
& you appear to weigh more.
F = m(g + a)
- If it accelerates downward, the scale expands
& you apparently weigh less.
F = m(g – a)
Ex2) Your mass is 75kg, you stand on a scale
in an elevator not moving.
a. What is your weight?
b. If the elevator accelerates UP at 2 m/s2 ,
what's your weight?
c. If it is moving at a constant speed
what's your weight?
d. Accelerates down at 2 m/s2?
e. Accelerates down at 5 m/s2?
f. Accelerates down at 9.8 m/s2?
Ex3) A 50kg bucket is lifted by a rope that has a max tension of 550N.
It started at rest, and after 3m is moving at 3 m/s,
will the rope break?
Ch6 HW#2 9 – 13
Lab6.2 – Newton’s 2nd Law of Skateboards
- due tomorrow
- Ch6 HW#2 due at beginning of period
Ch6 HW#2 9 – 13
9. Draw. A skydiver falls downward through the air
at a constant velocity.
10. Draw. A rope lifts a bucket accelerating it upward.
11. Draw. A rocket blasts off and its vertical velocity
increases with time. (No air drag.)
12. You weigh 585N on Earth.
a. Mass?
b. Weight on moon? (g=1.60 m/s2)
Fgm =
13. Mass of 60kg. What does scale read:
a. Elevator not moving?
Fg=
b. Up @ const speed?
Fg=
c. If the elevator were moving downward
and then began slowing to a stop, scale reads _____.
d. If the elevator were moving upward
and then began slowing to a stop, scale reads _____.
e. Slows at 2.0 m/s2 while moving upward?
f. Speeds up at 2.0 m/s2 while moving downward?
Ch6.3 – Friction
Kinetic Friction (Ff,k ) – moving friction
Ff,k = µk × FN.
Coefficient
Normal force
of kinetic friction (usually equal & opposite to Fg)
Ex1) Calculate the kinetic friction for the steel blades of ice skates
on ice given the total mass of the skater is 76 kg and µk =.06
Ch6.3 – Friction
Kinetic Friction (Ff,k ) – moving friction
Ff,k = µk × FN.
Coefficient
Normal force
of kinetic friction (usually equal & opposite to Fg)
Ex1) Calculate the kinetic friction for the steel blades of ice skates
on ice given the total mass of the skater is 76 kg and µk =.06
(Fg = (75kg.)(10 m/s2)= 750N)
FN
Ff,k
Fg
Ff,k= µk × FN
=(.06)Fg
=(.06)(750N)
= 45N
Ex2) What is the coefficient of friction for steel on steel without lubrication,
if the mass on the block on top is 10kg,
and the friction force is measured to be 30N.
Ex2) What is the coefficient of friction for steel on steel without lubrication,
if the mass on the block on top is 10kg,
and the friction force is measured to be 30N.
Ff,k = µk × FN
30N = µk(100N)
µk = .3
Static Friction- (Ff,s) Not moving friction
(prevents motion from happening)
Ff,s = µs × FN
Coefficient Normal force
of static friction
Static Friction can vary between 0 and this maximum value
Static Friction- (Ff,s) Not moving friction
(prevents motion from happening)
Friction depends on:
Ff,s = µs × FN
1. the types of surfaces (µ)
2. the force pressing the
Coefficient Normal force
surfaces together (FN)
of static friction
(not speed or surface area)
Static Friction can vary between 0 and this maximum value
Static Friction- (Ff,s) Not moving friction
(prevents motion from happening)
Friction depends on:
Ff,s = µs × FN
1. the types of surfaces (µ)
2. the force pressing the
Coefficient Normal force
surfaces together (FN)
of static friction
(not speed or surface area)
Static Friction can vary between 0 and this maximum value
Ex3) A person pushes on a wooden crate on a wood floor,
with a force of 50N. If the crate has a mass of 25kg,
and µs = 0.5, will the crate move?
Static Friction- (Ff,s) Not moving friction
(prevents motion from happening)
Friction depends on:
Ff,s = µs × FN
1. the types of surfaces (µ)
2. the force pressing the
Coefficient Normal force
surfaces together (FN)
of static friction
(not speed or surface area)
Static Friction can vary between 0 and this maximum value
Ex3) A person pushes on a wooden crate on a wood floor,
with a force of 50N. If the crate has a mass of 25kg,
and µs = 0.5, will the crate move?
FN
Ff,s = µs × FN
= (.5)(250N)
Ff,s
F = 50N
= 125N
Fg
Lab ex) Your group weighs a block with a spring scale
and finds it to be 15N. The force required to pull it
at constant speed is 12N.
What is the coefficient of friction between the block and table?
Lab ex) Your group weighs a block with a spring scale and finds it to be 15N.
The force required to pull it at constant speed is 12N.
What is the coefficient of friction between the block and table?
FN
Ff,k = µk × FN
12N = (µk)(15N)
Ff,s
F = 12N
µk = 0.8
Fg
Ch6 HW #3 14 – 17
Lab6.3A – Calculating Coefficients of Friction
- due tomorrow
- Ch6 HW#3 due at beginning of period
Ch6 HW#3 14 – 17
14. Calculate the kinetic friction for the rubber soled shoe on wet concrete.
mass of 55kg, µk = 0.40.
15.What is the coefficient of static friction for you (55kg) in your
rubber soled shoes if the friction force required to get you sliding is 350N.
Ch6 HW#3 14 – 17
14. Calculate the kinetic friction for the rubber soled shoe on wet concrete.
mass of 55kg, µk = 0.40.
FN
Ff,k = µk × FN
Ff,k = (.4)(550N)
Ff,k
F
Ff,k = 220N
Fg
15.What is the coefficient of static friction for you (55kg) in your
rubber soled shoes if the friction force required to get you sliding is 350N.
16. A person pushes on a wooden crate on a wood floor.
Crate has a mass of 45kg,
µs = 0.50, what force necessary to get it started moving?
FN
Ff,s = µs × FN
Ff,s
F
Fg
17. Your lab group weighs a block with a spring scale and finds
it weighs 9.0N. The force required to move it at constant speed
across light sand paper is 15.5N. What is µk?
FN
Ff,k
Ff,k = µk × FN
F
Fg
Ch6.3B – Friction cont
Ex1) A lab group pulls a 6kg ball at constant speed across a table
with a force of 15N. What’s the coefficient of friction?
FN
F= 15N
Ff,k
Fg
Ex1) A lab group pulls a 6kg ball at constant speed across a table
with a force of 15N. What’s the coefficient of friction?
FN
Fnet = F – Ff,k
F= 15N
Ff,k
m.a = 15N – μk ×FN
0 = 15 – μk ×60N
Fg
μk = .25
Ex2) Force of 2.5N is required to get a .4kg. block sliding.
A force of 2N is required to keep it moving at a constant speed.
Find µs and µk.
Ex2) Force of 2.5N is required to get a .4kg. block sliding.
A force of 2N is required to keep it moving at a constant speed.
Find µs and µk.
FN
FF,S
F = 2.5N
2.5N= μs × 4N
Fg= 4N
FF,K
FN
F = 2N
Fg= 4N
FF,S= μs×FN
μs= .63
FF,K= µk × FN
2N = µk × 4N
μ = .5
Ex3) A person pushes a crate of 20kg with a force of 101N.
μs = 0.50. Will the Crate move?
What force will be needed to keep it moving if μk = 0.20?
If the person still pushes with 101N,
what is the acceleration of the crate?
Ex3) A person pushes a crate of 20kg with a force of 101N.
μs = 0.50. Will the Crate move?
FN
Is F > Ff,s ?
Ff,s = μs x FN
Ff,s
F=101N
= (.5)(200N)
Fg=200N
= 100N
Yes!
What force will be needed to keep it moving if μk = 0.20?
If the person still pushes with 101N,
what is the acceleration of the crate?
Ex3) A person pushes a crate of 20kg with a force of 101N.
μs = 0.50. Will the Crate move?
FN
Is F > Ff,s ?
Ff,s = μs x FN
Ff,s
F=101N
= (.5)(200N)
Fg=200N
= 100N
Yes!
What force will be needed to keep it moving if μk = 0.20?
Fnet = F – Ff,k
0 = F – μk.FN
F = (.2)(200N)
= 40N
If the person still pushes with 101N,
what is the acceleration of the crate?
Ex3) A person pushes a crate of 20kg with a force of 101N.
μs = 0.50. Will the Crate move?
FN
Is F > Ff,s ?
Ff,s = μs x FN
Ff,s
F=101N
= (.5)(200N)
Fg=200N
= 100N
Yes!
What force will be needed to keep it moving if μk = 0.20?
Fnet = F – Ff,k
0 = F – μk.FN
F = (.2)(200N)
= 40N
If the person still pushes with 101N,
what is the acceleration of the crate?
Fnet = F – Ff,k
m.a = 101N – 40N
(20kg).a = 61N
a = 3.1 m/s2
Ch6 HW#4 18,19
Lab6.3B – Sliding Friction
- due tomorrow
- Ch6 HW#4 due at beginning of period
Ch6 HW#4 18,19
18. A person pushes a crate of 30kg with a force of 25N.
μs = 0.78. Will the Crate move?
Is F > Ff,s ?
Will the crate move if μs = 0.04?
Ff,s = μs x FN
If the person still pushes with 25N,
what is the acceleration of the crate? (μk = 0.04 also)
Fnet = F – Ff,k
19. A lab group pulls on a 500g cart with a spring scale that reads 4N.
The coefficient of static friction, μs, is 0.30. Will the cart move?
FN
Is F > Ff,s ?
Ff,s = μs x FN
Ff,s
F=4N
Fg=5N
The lab group continues to pull with a force of 4N.
the coefficient of kinetic friction is 0.10.
What is the acceleration of the cart?
Fnet = F – Ff,k
m.a = 4N – μk x FN
(.5kg).a = 4N – (0.1)x(5N)
Ch6.3C – Friction cont
Lab6.3B Data Analysis
Data Table
Total Mass of Block
Total Mass of Hangar
1. Just Friction block
2. Block + 200g
3. Block + Unknown
---------------------------------
Calculations Table
Total Weight
of Block
Total Weight
of Hangar
Coefficient of
friction, μ
1. Block
2. Block+200g
3. Block+ Unknown
-------------------------
--------------------
Calculations Table
Total Weight
of Block
Total Weight
of Hangar
Coefficient of
friction, μ
1. Block
2. Block+200g
3. Block+ Unknown
-------------------------
Solving for the unknown block:
1. Average μ:
2. Solve for FN: Ff = μ.FN
3. FN = Fg subtract out block.
4. F = m.g, solve for mass:
--------------------
Ex1) Using the apparatus from Lab6.3B, what mass would pull
a 250g friction block with a coefficient of kinetic friction, μk, of 0.55?
Lab ex) A 0.25kg ball rolls off a ramp and down a piece of carpet ,
and rolls 5.5m during a time of 3.0sec.
Find the friction between the carpet and ball.
1.
2.
3.
4.
a = _____m/s2
vi =___m/s d = 5.5m
t = 3 sec.
vf = 0
Lab ex) A 0.25kg ball rolls off a ramp and down a piece of carpet ,
and rolls 5.5m during a time of 3.0sec.
Find the friction between the carpet and ball.
1. d = ½(vi+vf).t
2. vf = vi + at
3. F = m.a
vi = 3.7m/s
5.5 = ½(vi + 0)3
0 m/s = 3.7m/s + a(3s)
a = - 1.23m/s2
F = (0.25kg)(-1.23m/s2)
F = - 0.3N
0.3N = μk x (2.5N)
μk = 0.12
4. Ff,k = μk x FN
a = - 1.23 m/s2
vi = 3.7m/s d = 5.5m
t = 3 sec.
Ch6 HW#4B 1,2
vf = 0
FN
FF,k
Fg
Lab6.3C – Friction on Carpet
- due tomorrow
Ch6 HW#4B 1,2
1. A 14kg ball rolls off a ramp and down a piece of carpet ,
and rolls 25.2m during a time of 4.5sec.
Find the friction between the carpet and ball.
1. d = ½(vi+vf).t
2. vf = vi + at
3. F = m.a
4. Ff,k = μk x FN
a = ___
m/ 2
s
d= 25.2m vf= 0
t= 4.5 sec.
FN
FF,k
Fg
2. Using the apparatus from Lab6.3B, what mass would pull
a 1000g friction block with a coefficient of kinetic friction, μk, of 0.80?
Ch6.4 – Periodic Motion
Simple harmonic motion – An object oscillates back and forth.
Period - how long it takes to complete one cycle
Amplitude - Its maximum distance from rest position.
Equilibrium - rest position, all forces cancelling out.
1. A Pendulum
Ch6.4 – Periodic Motion
Simple harmonic motion – An object oscillates back and forth.
Period - how long it takes to complete one cycle
Amplitude - Its maximum distance from rest position.
Equilibrium - rest position, all forces cancelling out.
1. A Pendulum
FT
FT
FNet
Fg
FT
Fg
Period of a pendulum:
Ch6.4 – Periodic Motion
Simple harmonic motion – An object oscillates back and forth.
Period - how long it takes to complete one cycle
Amplitude - Its maximum distance from rest position.
Equilibrium - rest position, all forces cancelling out.
1. A Pendulum
FT
FT
FNet
Fg
FT
Fg
Depends on string length
Period of a pendulum:
and accl of gravity.
Pendulums don’t care about the mass attached!
Ex1) What is the length of a pendulum with a period of 2 seconds?
Ex1) What is the length of a pendulum with a period of 2 seconds?
l
T p  2
g
l
2  2
9.8
l  .99m
Ch6 HW#5 20 - 23
Ch6 Lab Quiz
- Pendulums
Ch6 HW#5 20 – 23
20. Period of a pendulum with length of 0.10m?
21. Length of pendulum with period of 5s?
22. Accl of gravity on planet where pendulum
of 1m length has a period of 1 sec?
23. Length of a pendulum with a period of 10s?
Ch6.4B – Periodic Motion – Springs
Springs – amount they stretch is proportional to the mass attached.
F = k.d
elastic spring distance
force
constant displaced
Ex1) A 100g mass is attached to a spring and it stretches 10 cm.
What is the value of the spring constant?
Period of a Spring
m
Ts  2
k
springs care about the mass attached
and the stretchiness of the spring
to determine the time to oscillate.
Ex3) What is the spring constant for a spring that has a period of 0.63s
when a mass of 100g is attached?
Period of a Spring
m
Ts  2
k
springs care about the mass attached
and the stretchiness of the spring
to determine the time to oscillate.
Ex3) What is the spring constant for a spring that has a period of 0.63s
when a mass of 100g is attached?
m
Ts  2
k
1N
0.63s  2
k
k  10 N
m
Resonance – with a small
force applied, an object will
oscillate at its natural frequency.
Keep applying force, amplitude
increases tremendously.
Ch6 HW#6 24 – 27
Lab6.4 – Periodic Motion
- due tomorrow
- Ch6 HW#6 due at beginning of period
Ch6 HW#6 24 – 27
24. Mass of 0.250kg suspended from a spring
that stretches 0.05m. Find spring constant.
25. A 0.01kg mass suspended from a spring with
a spring constant of 10 N/m. How much stretch?
26. A spring with a spring constant of 5 N/m has a 50g mass attached,
and is pulled down and released,
setting it into simple harmonic motion. What is its period?
27. A spring with a spring constant of 20 N/m has a mass attached,
and is pulled down and released,
setting it into simple harmonic motion. Its period is 0.63 sec.
What mass is attached?
Ch6.5 – Newton’s 3rd Law
- For every action (force) there is an equal and opposite reaction (force)
- Forces always come in pairs (Interaction pairs)
FA on B = FB on A
Ex1) A 100kg football player pushes off his 70kg girlfriend while ice skating.
If her acceleration is 6 m/s2, what is his?
Ex2) A 50kg person jumps upwards off Earth’s surface w/ an accl of 3 m/s2.
What is the acceleration of the Earth downward? (ME= 6×1024 kg)
Ex3) A 300kg astronaut is skydiving on the moon. If he free falls at 1.6 m/s2,
how fast does the moon accelerate up? (Mm = 7×1022 kg)
Ex4) A 1000kg truck starts from rest and after 3 sec is travelling 12 m/s.
By how much does the earth accelerate its spin?
FT on E
FE on T
Ex4) A 1000kg truck starts from rest and after 3 sec is travelling 12 m/s.
By how much does the earth accelerate its spin?
FT on E
FE on T
vf = vi + at
12m/s = 0 + a(3)
a = 4 m/s2
FE on T = FT on E
mT×a = mE×a
(1000)(4) = (6×1024)(a)
a = 7×10-22 m/s2
Ch6 HW#7 28 – 31
Ch6 HW#7 28 – 31
28. 7kg rock free falls, what is accl of earth?
29. 8000kg truck brakes on 50,000kg asteroid with a deceleration
of 20 m/s2, what is accl of asteroid?
30. 65kg boy in tug-o-war with 45kg girl, she accl at 3 m/s2,
what is his accl?
FG on B = FB on G
mB×a = mG×a
31. What is the force on the boy?
vi = 30m/s
vf = 0m/s
t = 0.005s
a=?
vf = vi + at
0 = 30 + a(0.005)
a = -6000 m/s2
FGlove on Ball = FBall on Glove
mB×a = mG×a
Ch6.6 – Coupled Motion
Ex1) Find the acceleration of this system:
..
1kg
5kg
Ex2) Find the acceleration of this system:
..
2kg
4kg
Ex3) Find the acceleration of this system:
..
10kg
3kg
Ch6 HW#8 32 – 35
Lab6.5 – Coupled Motion
- due tomorrow
- Ch6 HW#7 due at beginning of period
Ch6 HW#8 32 – 35
32) Find the acceleration of this system:
..
Fnet = Fg3 – FT + FT – Fg2
2
3
33) Find the acceleration of this system:
..
Fnet = Fg5 – FT + FT – Fg3
3
5
34) Find the acceleration of this system:
..
Fnet = Fg5 – FT + FT – Fg1
1
5
35) Find the acceleration of this system:
..
Fnet = Fg3 – FT + FT – Fg3
3
3
Ch6 HW#9 36 - 38
36) Find the acceleration of this system:
2
.
.
3
Fnet = Fg3 – FT + FT
37) Find the acceleration of this system:
3
.
.
5
Fnet = Fg5 – FT + FT
38) Find the acceleration of this system:
5
.
.
1
Fnet = Fg1 – FT + FT
Ch6 Rev
1. A rock is dropped from a bridge into a valley. The Earth pulls on the rock
and accl it downward. According to Newton’s 3rd Law, the rock must
also be pulling on the Earth, yet the Earth doesn’t seam to accl. Explain.
2. State Newton’s 3 Laws.
3. A 873kg dragster starting from rest attains a speed of 26.3 m/s in 0.59s.
a. Find accl
vf = vi + at
b. Find net force
F = m.a
4. A bowling ball has a mass of 8.0kg. What is its weight?
Ch6 Rev
5. Your new motorcycle weighs 2450N. What is its mass?
6. A pendulum has a length of 0.67m. Find its period.
7. What is the length of a pendulum with a period of 2.5s?
Ch6 Rev
8. What is the mass of an object hanging from a spring if the period
of oscillation is 3s, and the spring constant has a value of 50 N/m.
9. If a mass weighing 20N is suspended from a spring causing it to stretch
15cm, what is the value of the spring constant?
Ch6 Rev
10. If you use a horizontal force of 30N to pull a 12.0kg wooden crate across
a floor at constant velocity, what is the coefficient of kinetic friction
between the crate and floor?
Ff,k = μk x FN
FN
Ff,k
F=12N
Fg=120N
11. Calculate the kinetic friction force when a 200kg sled slides across
a frozen lake, coefficient of kinetic friction = 0.06.
Ch6 Rev
12. A person pushes a 25kg box with a force of 26N, coefficient of static
friction is 0.43.
a. Will the crate move?
Ff,s = μs x FN
FN
Ff,s
F=26N
Fg=250N
b. If coefficient decreases to 0.10, will it move?
c. If the person still pulls with 26N, and the coefficient of kinetic
friction is 0.05, what is the accl?
Ch6 Rev
13. A 60kg skydiver jumps from an airplane.
a. What is the force of the earth on skydiver?
b. What is the force of the skydiver on earth?
c. If the mass of the earth is 6x1024kg, what is its accl?
Ch6 Rev
14. What is the accl of this system?
..
15. What is the accl of this system?
..
Fnet = Fg3 – FT + FT – Fg1
1kg
2kg
3kg
7kg