Download Newtons Lesson 10

NEWTON’S LESSON 10
ELEVATOR PHYSICS
What are the forces at play on our bodies during the ride?
During the upward acceleration?
Normal force exceeds weight, and the body feels heavy because the floor of
the elevator is pushing us up harder than usual.
During constant velocity between floors?
No net force on our bodies and the riders feel relatively normal.
During downward acceleration?
Normal force is less than our weight, and the body feels light because the
floor is not pushing us up as hard as it ordinarily does and there is less force
on the soles of our feet.
What you feel, we call it your apparent weight, is not the pull of gravity
down on you, rather what you feel as your weight is the floor pushing up on
you.
• In other words, the apparent weight you feel is actually the normal force of
the elevator floor acting on you (Fapparent = FN).
• The normal force changes due to
the motion of the elevator floor.
If you are accelerating up (Fnet > 0) your Apparent weight will be greater.
(Examples????: Accelerating as you go from rest and ascend,
decelerating as you come to a stop at the ground floor)
If you are accelerating down (Fnet < 0) your apparent weight will be
smaller.
(Examples????: Decelerating as you come to a stop at top of the
ride, accelerating as you go from rest and descend down)
When you are at rest or moving at constant speed (Fnet =0) your apparent
weight will be your normal weight.
Example:
If you normally weigh 706 N, what is your apparent weight if you are in an
elevator that is slowing down at the rate of 1.65 m/s2 ?
First, draw a free body diagram and determine the direction of acceleration:
Acceleration is downward as the direction of motion is up but the object is
slowing down and coming to a rest, therefore there is an acceleration
downward.
Second, determine the mass of the object using the Fg:
Fg = mg
m = Fg/g = 706/9.81 = 71.9674 kg
Third, determine an equation that represents the forces acting on you:
Fnet = Fg - FN
71.9674*(1.65) = 706 – FN
FN = 706 - 118.7462 = 587 N, up
Example:
A lamp hangs vertically from a cord in an elevator which is descending with
an downward acceleration of a=2.00m/s2. The tension in the cord is
FT=10.0N. What is the mass m of this lamp?
First, draw a free body diagram and determine the direction of acceleration:
Acceleration is downward.
Second, determine an equation that represents the forces acting on the lamp:
Fnet = Fg – FT
m*(2.00) = m*(9.81) – 10.0
10.0= m(9.81 – 2.00)
m = 1.28 kg
Example:
A 1200.0 kg rocket produces 30000.0 N of thrust. What is the speed of the
rocket 45.0 s after launch?
First, draw a free body diagram and determine the direction of acceleration:
Acceleration is upward.
Second, determine an equation that represents the forces acting on the rocket
and find the acceleration of the rocket:
Fnet = Fthrust – Fg
1200(a) = 30000 - 1200*(9.81)
a = 15.19 m/s2 , upward
Third, using the acceleration, find the final velocity of the rocket using the
equation:
vf = vi + at
vf = 0 + 15.19(45)
vf = 684 m/s, upward
NEWTON’S LESSON 10 HOMEWORK
1.
2.
3.
4.
What is the acceleration of an elevator at rest?
What is the net force when the elevator travels at a constant velocity?
When do most elevators accelerate during their motion?
When is the net force on the elevator not 0.0 N?
Elevator at Rest or Moving at Constant Velocity:
5. When a person stands on a bathroom scale while on the elevator, what is his weight
when the elevator is at rest?
6. Moving at constant speed?
7. What is the net force acting on the person?
8. What is the acceleration of the person?
Elevator Accelerating Upward:
9. When the elevator accelerates upward the person must accelerate upward.
Which force must increase the weight of the person or the normal force?
10. What does the elevator passenger feel during this upward acceleration?
11. Why can't the weight of the person decrease?
Elevator Accelerating downward:
12. When the elevator accelerates downward the person must accelerate downward.
Which force must decrease the weight of the person or the normal force?
13. What does the elevator passenger feel during this downward acceleration?
14. When would the passenger feel weightless on an elevator?
15. Why does the passenger feel weighless? (Hint: what is the value of the normal force?)
16. What happens to the passenger when the elevator accelerates downward at a value
greater than -9.81 m/s2?
17. What is the source of the upward force on the passenger in an elevator?
18. Is our sense of weight determined by the pull of gravity on us or the normal force
coming from the floor or the ground we stand upon?
19. You are standing on a scale in an elevator. You have a mass of 75kg. Determine
what a scale would show as your “apparent” mass (in kilograms) if…
a. the elevator starts to accelerate upwards at 3.0m/s2.
b. the elevator starts to accelerate downwards at 4.0m/s2
HINT: Having trouble converting FN into kg? When was the last time you saw a
bathroom scale give a reading in Newtons? You can change this into a reading in
kilograms by remembering that the scale we're using has no idea what is going on... it
still thinks it's sitting in someone's bathroom where gravity is a nice constant 9.81m/s2.
Therefore, FN = m(g) for this situation.
20. A jet accelerates vertically up at 8.5 m/s2. What force does the exhaust gas exert on
the 4400 kg jet?
21. A 2.00 kg pendulum hangs in an elevator. Calculate the tension in the string
supporting the pendulum if the elevator moves:
a. with zero velocity
b. downward at a constant velocity of 2.5 m/s
c. upward at a constant velocity of 2.5 m/s
d. downward at a constant acceleration of 2.00 m/s2
e. upward at a constant acceleration of 2.00 m/s2
22. A man measures the acceleration of an elevator by using a spring balance. He fastens
the scale to the roof, and suspends a mass from it. The scale reads 98 N when the elevator
is at
rest, and 93 N when the elevator is moving.
a. In which direction is the elevator accelerating?
b. What is the acceleration of the elevator?
23. A fish hangs from a spring scale supported from the roof of an elevator. If the
elevator has an upward acceleration of 1.2 m/s2 and the scale reads 200.0 N, what is the
true force of gravity on the fish? (b) Under what circumstances will the scale read 150 N
and what would be the acceleration of the elevator? (c) What will the scale read if the
elevator cable breaks?
24. A 0.500 kg soccer ball is kicked from the top of a 14.0-meter cliff with a horizontal
speed of 28.0 m/s. What distance from the bottom of the cliff does the ball land? Please
show your work in an organized fashion (listing known and unknown values) (23.7 m)
HOMEWORK KEY FOR NEWTON’S LESSON 10 HOMEWORK
1. 0 m/s2
2. Fnet =0
3. start/stop
4. starts/stops
5. same as his actual weight
6. actual weight
7. Fnet = 0
8. a = 0
9. Fnet
10.heavier
11.Fnet is in the direction of FN (apparent weight)
12.Fnet
13.lighter
14.if elevator was in free fall
15.no FN pushing up on passenger
16.passenger is in free fall and is weightless, no FN
17.FN
18.FN
19.a. 98 kg, b. 44 kg
20. 8.1 x 104 N
21. a. 19.6 N, b. 19.6 N, c. 19.6 N, d. 15.6 N, e. 23.6 N
22. a. down, b. 0.51 m/s2
23.a. 1.8 x 102 N, b. either going up and slowing down or going down
and speeding up. , 1.6 m/s2 down, c. FT = 0
24. 23.7 m
ELEVATOR WORKSHEET
The "Elevator Problem" is a classic problem in
physics. The situation is this:
"You are standing on a bathroom scale in an
elevator. You are holding an apple. (Yes,
people are staring at you...) You weigh 500
Newtons, so your mass is about 50 kg."
This assignment is a step-by-step analysis of the
elevator problem. A good deal of the work has been
done for you - in which case it is your job to study the answers given and use them as a
pattern and guide for the answers that you supply.
Part A: Elevator Is At Rest.
You have just boarded the elevator, so it (with you inside) is at rest...
What does the scale read?
Answer:
There are 2 forces acting on you. (See the
diagram at right.) The Earth pulls down on
you with the force we call your weight (= mg)
of 500 Newtons.
Since the elevator is at rest, your acceleration is 0 m/s2. Since your acceleration is 0 m/s2,
Newton's First Law says the net force on you must be 0 Newtons.
Since the net force on you is 0 Newtons, the upward forces and downward forces on you
must balance exactly. Therefore the scale must push on you with a force of (1a) _______
Newtons, and the scale must read (1b) _______ Newtons.
Question 2: If you let go of the apple, what does it do?
Answer:
The apple would be in free fall, so its acceleration relative to the earth is 10 m/s2
downward. Since you are at rest relative to the earth, the apple's acceleration relative to
you would be10 m/s2 also, so the apple would appear to fall just as it does anywhere else
on the earth.
Part B: The Elevator Accelerates Upward.
The elevator, (with you inside) begins
to accelerate upward from rest at 2
m/s2.
Question 3: What will the scale
read now?
Answer:
There are 2 forces acting on you.
(Complete the diagram.) Your weight
pulls down with a force of 500
Newtons. The scale pushes up with a
force of (3a) ______ (see below) Newtons.
Since your acceleration is 2 m/s2 upward, Newton's Second Law says that there must be a
net force pushing you upward, and the net force has a magnitude Fnet = ma. So the net
force on you,
Fnet = (50 kg)(2 m/s2) = 100 Newtons (upward).
Since the net force on you is100 Newtons, the upward forces and downward forces on
you must cancel to leave a 100 Newton upward force. Therefore the scale must push on
you with a force of (3b) _______ Newtons, and the scale must read (3c) _______
Newtons as the elevator accelerates upward.
Question 4: If you let go of the apple now, what does it do?
Answer:
The apple would be in free fall, so its acceleration relative to the earth is 10 m/s2
downward. Since you are accelerating at 2 m/s2 upward relative to the earth, the apple's
acceleration relative to you would be 10 m/s2 + 2 m/s2 = 12 m/s2, so the apple would
appear to fall faster inside the elevator than it does in "normal" free fall on the earth.
To the occupants of the upwardly accelerating elevator, it appears that gravity is
stronger, since they seem to weigh more (why?) and objects fall faster than "normal".
Part C: The Elevator Moves Up With Constant Velocity.
The elevator (and you) accelerated for 5 seconds, so it is moving
upward with a velocity of 10 m/s. It now moves with this
constant upward velocity of 10 m/s.
Question 5: What does the scale read now?
Answer:
There are 2 forces acting on you. (Complete the diagram.) Your
weight pulls down with a force of 500 Newtons. The scale
pushes up with a force of (5a) ________ (see below) Newtons.
Since the elevator is moving with constant velocity, your acceleration is (5b) ____ m/s2.
Since your acceleration is (5c) ____ m/s2, Newton's First Law says the net force on you is
(5d) _____ Newtons.
Since the net force on you is (5e) ____ Newtons, the scale must push on you with a force
of (5f) _______ Newtons, and the scale must read (5g) _______ Newtons.
Question 6: If you let go of the apple, what does it do?
Part D: The Elevator Slows Down (While Going Up)
The elevator, (with you inside) begins to slow down as it
approaches its destination. Its acceleration (or deceleration)
is 2 m/s2 downward.
Question 7: What does the scale read now?
Answer:
There are 2 forces acting on you. (Complete the diagram.)
The Earth (your weight) pulls down with a force of (7a)
_____ Newtons. The scale pushes up with a force of
(7b)______ (see below) Newtons.
Since your acceleration is 2 m/s2 downward, Newton's Second Law says that there must
be a net force pulling you downward, and the net force has a magnitude Fnet = ma. So the
net force on you,
Fnet = ((7c) _____ kg)((7d) _____ m/s2) = (7e) ______ Newtons .
Since the net force on you is (7f) _____ Newtons downward, the upward forces and
downward forces on you must cancel out to leave a (7g) _____ Newton downward force .
Therefore the scale must push on you with a force of (7h) _______ Newtons, and the
scale must read (7i) _______ Newtons as the elevator accelerates downward.
Question #8: If you let go of the apple now, what does it do?
To the occupants of the downwardly accelerating elevator, it appears that gravity is
weaker, since they seem to weigh less (why?) and objects fall more slowly than "normal".
Part E: The Elevator Speeds Up (While Going Down)
The elevator, (with you inside) reaches its floor, stops for a while, and then begins to
accelerate downward. Its acceleration is 2 m/s2 downward.
Question 9: What does the scale read now?
Answer:
There are 2 forces acting on you. (Complete the diagram.) The
Earth (your weight) pulls down with a force of (9a) _____
Newtons. The scale pushes up with a force of (9b)______ (see
below) Newtons.
Since your acceleration is 2 m/s2 downward, Newton's Second
Law says that there must be a net force pulling you downward,
and the net force has a magnitude Fnet = ma. So the net force on
you,
Fnet = ((9c) _____ kg)((9d) _____ m/s2) = (9e) ______ Newtons .
Since the net force on you is (9f) _____ Newtons downward, the upward forces and
downward forces on you must cancel out to leave a (9g) _____ Newton downward force.
Therefore the scale must push on you with a force of (9h) _______ Newtons, and the
scale must read (9i) _______ Newtons as the elevator accelerates downward.
Question #10: If you let go of the apple now, what does it do?
To the occupants of the downwardly accelerating elevator, it appears that gravity is
weaker, since they seem to weigh less (why?) and objects fall more slowly than "normal".
Part F: The Elevator Moves Down With Constant Velocity.
The elevator (and you) accelerated for 5 seconds, so it is moving downward with a
velocity of 10 m/s. It now moves with this constant
downward velocity of 10 m/s.
Question 11: What does the scale read now?
Answer:
There are 2 forces acting on you. (Complete the
diagram.) Your weight pulls down with a force of 500
Newtons. The scale pushes up with a force of (11a)
________ (see below) Newtons.
Since the elevator is moving with constant velocity, your acceleration is (11b) ____ m/s2.
Since your acceleration is (11c) ____ m/s2, Newton's First Law says the net force on you
is (11d) _____ Newtons.
Since the net force on you is (11e) ____ Newtons, the scale must push on you with a
force of (5f) _______ Newtons, and the scale must read (11g) _______ Newtons.
Question 12: If you let go of the apple, what does it do?
Part G: Oh, No!
The elevator cable snaps, and the elevator (with you inside!) begins to fall! Perhaps you
have time for one last Physics observation!
Question 13: What does the scale read as the elevator falls?
Answer:
There are 2 forces acting on you. (Complete the
diagram.) The Earth (your weight) pulls down with a
force of (13a) _____ Newtons. The scale pushes up
with a force of (13b) _____ (see below) Newtons.
Since the elevator and you are in free fall, your
acceleration is (13c) ____ m/s2 downward. Newton's
Second Law says that there must be a net force
pulling you downward, and the net force has a
magnitude Fnet = ma. So the net force on you,
Fnet = ma = ((13d) _____ kg)((13e) _____ m/s2) = (13f) _____ Newtons.
Since the net force on you = (13g) _____ Newtons downward, the upward forces and
downward forces on you must cancel out to leave a (13h) _____ Newton downward force
. Therefore the scale must push on you with a force of (13i) _______ Newtons, and the
scale must read (13j) _______ Newtons as the elevator accelerates downward.
Question #14: If you let go of the apple now, what does it do?
Question #15: What can you do to be weightless in the elevator?
NEWTON’S LESSON 10 HOMEWORK
1.
2.
3.
4.
What is the acceleration of an elevator at rest?
What is the net force when the elevator travels at a constant velocity?
When do most elevators accelerate during their motion?
When is the net force on the elevator not 0.0 N?
Elevator at Rest or Moving at Constant Velocity:
5. When a person stands on a bathroom scale while on the elevator, what is his weight
when the elevator is at rest?
6. Moving at constant speed?
7. What is the net force acting on the person?
8. What is the acceleration of the person?
Elevator Accelerating Upward:
9. When the elevator accelerates upward the person must accelerate upward.
Which force must increase the weight of the person or the normal force?
10. What does the elevator passenger feel during this upward acceleration?
11. Why can't the weight of the person decrease?
Elevator Accelerating downward:
12. When the elevator accelerates downward the person must accelerate downward.
Which force must decrease the weight of the person or the normal force?
13. What does the elevator passenger feel during this downward acceleration?
14. When would the passenger feel weightless on an elevator?
15. Why does the passenger feel weighless? (Hint: what is the value of the normal force?)
16. What happens to the passenger when the elevator accelerates downward at a value
greater than -9.81 m/s2?
17. What is the source of the upward force on the passenger in an elevator?
18. Is our sense of weight determined by the pull of gravity on us or the normal force
coming from the floor or the ground we stand upon?
19. You are standing on a scale in an elevator. You have a mass of 75kg. Determine
what a scale would show as your “apparent” mass (in kilograms) if…
a. the elevator starts to accelerate upwards at 3.0m/s2.
b. the elevator starts to accelerate downwards at 4.0m/s2
HINT: Having trouble converting FN into kg? When was the last time you saw a
bathroom scale give a reading in Newtons? You can change this into a reading in
kilograms by remembering that the scale we're using has no idea what is going on... it
still thinks it's sitting in someone's bathroom where gravity is a nice constant 9.81m/s2.
Therefore, FN = m(g) for this situation.
Answer:
a. First, draw a free body diagram:
acceleration is upwards, FN is upwards (apparent mass), Fg is downwards
Second, determine an equation that represents the forces acting on the person:
Fnet = FN – Fg
75(3.0) = FN – 75(9.81)
FN = 225 + 735.75 = 960.75 N
FN = mg
m = FN / g
= (960.75) / (9.81m/s2)
m = 98 kg
So a regular scale thinks you weigh 98 kg!
b. First, draw a free body diagram:
acceleration is downwards, FN is upwards (apparent mass), Fg is downwards
Second, determine an equation that represents the forces acting on the person:
Fnet = Fg - FN
75(4.0) = 75(9.81) - FN
FN = 735.75 - 300 = 435.75 N
FN = mg
m = FN / g
= (435.75) / (9.81m/s2)
m = 44 kg
So a regular scale thinks you weigh 44 kg!
20. A jet accelerates vertically up at 8.5 m/s2. What force does the exhaust gas exert on
the 4400 kg jet?
Answer: First, draw a free body diagram and determine the direction of acceleration:
Acceleration is upward.
Second, determine an equation that represents the forces acting on the rocket and find the
thrust from the exhaust gas:
Fnet = Fthrust – Fg
4400(8.5) = Fthrust - 4400*(9.81)
Fthrust = 37400 + 43164 = 80564 = 8.1 x 104 N
21. A 2.00 kg pendulum hangs in an elevator. Calculate the tension in the string
supporting the pendulum if the elevator moves:
a. with zero velocity
b. downward at a constant velocity of 2.5 m/s
c. upward at a constant velocity of 2.5 m/s
d. downward at a constant acceleration of 2.00 m/s2
e. upward at a constant acceleration of 2.00 m/s2
Answer:
a. zero velocity means there is no Fnet therefore tension in the string would equal the
weight of the pendulum Fg = FT = mg = 2.00*9.81 = 19.62 = 19.6 N
b. constant velocity means there is no acceleration and therefore no Fnet therefore the
tension in the string would equal the weight of the pendulum = 19.6 N
c. same as b.
d. First, draw a free body diagram:
Second, determine an equation that represents the forces acting on the pendulum and
determine the FT:
Fnet = Fg - FT
2.00(2.00) = 2.00*(9.81) - FT
FT= 19.62 - 4.0 = 15.62 = 15.6 N
e. First, draw a free body diagram:
Second, determine an equation that represents the forces acting on the pendulum and
determine the FT:
Fnet = FT - Fg
2.00(2.00) = FT - 2.00*(9.81)
FT= 19.62 + 4.0 = 23.62 = 23.6 N
22. A man measures the acceleration of an elevator by using a spring balance. He fastens
the scale to the roof, and suspends a mass from it. The scale reads 98 N when the elevator
is at
rest, and 93 N when the elevator is moving.
a. In which direction is the elevator accelerating?
b. What is the acceleration of the elevator?
Answer:
First, draw a free body diagram and determine the direction of Fnet on the elevator:
At resting position FT = Fg . However, as the elevator moves(accelerates)
FT < Fg. This means that the object is lighter than at rest, therefore the elevator and
object must be accelerating downwards, therefore Fnet must be downwards.
Mass of the mass suspended from the roof can be determined from the FT at rest which is
equal to Fg = mg, 98/9.81 = m = 9.99 kg
Second, determine an equation that represents the forces acting on the pendulum and
determine the acceleration:
Fnet = Fg - FT
9.99(a) = 98 – 93
a = 0.51 m/s2
23. A fish hangs from a spring scale supported from the roof of an elevator. If the
elevator has an upward acceleration of 1.2 m/s2 and the scale reads 200.0 N, what is the
true force of gravity on the fish? (b) Under what circumstances will the scale read 150 N
and what would be the acceleration of the elevator? (c) What will the scale read if the
elevator cable breaks? (1.8 x 102 N, 1.5 m/s2, ?)
Answer:
a. First, draw a free body diagram and determine the direction of Fnet on the elevator:
Second, determine an equation that represents the forces acting on the fish and determine
the Fg:
Fnet = FT - Fg
m(1.2) = 200 – m(9.81)
m(1.2 + 9.81) = 200
m(11.01) = 200
m = 18.165
Fg = mg
Fg = 18.165(9.81) = 178.2 = 1.8 x 102 N
b. First, draw a free body diagram and determine the direction of Fnet on the elevator:
FT = 150 N is lighter than the Fg of 180 N, therefore the Fnet must be downwards.
The elevator must be either going up and slowing down or coming down and speeding
up.
Fnet = Fg - FT
18.165(a) = 178.2 – 150
a = 1.6 m/s2
c. Fnet = Fg in freefall and therefore FT = 0. The acceleration of the fish will be the same
as the acceleration of gravity, therefore FT =0, Fnet = Fg – FT.
The fish is in a state of freefall and is therefore weightless, does not have a FT pulling or
pushing upwards on it.
24. A 0.500 kg soccer ball is kicked from the top of a 14.0-meter cliff with a horizontal
speed of 28.0 m/s. What distance from the bottom of the cliff does the ball land? Please
show your work in an organized fashion (listing known and unknown values) (23.7 m)
Answer:
dxf = ???
diy = 14.0 m
vix = 28.0 m/s
viy = 0.0 m/s (its launched horizontally)
ax = 0 m/s/s (true for all projectiles)
ay = -9.80 m/s/s (true for all projectiles)
Since three pieces of vertical-information are now known, a vertical-equation can be
employed to find the time.
dfy – diy = viy • t + 0.5 • ay • t2
0 – 14.0 = 0 + 0.5(-9.8)(t2)
t = 0.845
Now the t value can be combined with the vix and ax value and used in an horizontalequation
dfx – dix = vix • t
dfx – 0 = 28.0(0.845)
dfx = 23.66 = 23.7 m
ANSWER KEY FOR ELEVATOR WORKSHEET
A. Elevator At Rest
1. What does the scale say?
a. scale pushes with a force of 500 Newtons (to balance the Earth's
pull)
b. scale says 500 Newtons (equal to the scale's push)
2. What does the apple do?
B. Elevator Accelerates Up
3. What does the scale read? To complete the force diagram you need to an
upward force on you due to the scale. This force turns out to be 600
Newtons - see below.
a. the scale pushes with a force of 600 Newtons, since when the
scale's force is combined with the Earth's force (weight = 500
Newtons) there has to be a 100 Newton force (upward) left over the net force.
b. 600 Newtons (same as part a)
c. 600 Newtons (same as the force the scale exerts)
4. What does the apple do?
The apple would be in free fall, so its acceleration relative to the earth is
10 m/s2 downward. Since you are accelerating at 2 m/s2 upward relative to
the earth, the apple's acceleration relative to you would be 10 m/s2 + 2
m/s2 = 12 m/s2, so the apple would appear to fall faster inside the elevator
than it does in "normal" free fall on the earth.
To the occupants of the upwardly accelerating elevator, it
appears that gravity is stronger, since they seem to weigh more
(why?) and objects fall faster than "normal".
C. Elevator at Constant Velocity
6. What does the scale read? To complete the force diagram, you need to add
an upward force on you due to the scale. This force turns out to be 500
Newtons - see below.
a.
b.
c.
d.
e.
f.
The scale's force on you is 500 Newtons. (see below)
Your acceleration is 0 m/s2 (since your velocity is constant)
0 m/s2 (same as part b)
The net force must be 0 Newtons (if the acceleration is 0 m/s2).
0 Newtons (same as part d)
The scale's force must be 500 Newtons since it must exactly
balance your weight (500 Newtons) so that the net force is 0
Newtons.
g. The scale reads 500 N (same as the scale's force).
7. What does the apple do if you let go? The apple would be in free fall, so its
acceleration relative to the Earth would be 9.8 m/s2. Its acceleration relative to
you would be the same (since your acceleration is 0 m/s2) so the apple would
fall exactly as it would if the elevator were at rest.
D. Elevator Accelerates Downward (While Going Up)
8. What does the scale read? To complete the diagram, add an upward force
on you due to the scale. This force turns out to be 400 Newtons - see below.
h.
i.
j.
k.
l.
m.
n.
o.
Your weight is 500 Newtons (given in the problem)
The scale's force is 400 Newtons (see below)
Your mass is 50 kilograms (given in the problem)
Your acceleration is 2 m/s2
The net force is 100 Newtons (= (50 kg)(2 m/s2))
100 N (same as part e)
100 N (same as part e)
The scale's force is 400 Newtons, since it must combine with your
weight (500 Newtons) to leave a net downward force of 100
Newtons.
p. The scale reads 400 Newtons (same as scale's force)
9. What does the apple do if you drop it? Relative to the Earth, the apple
accelerates toward the Earth at about 10 m/s2, but you are accelerating toward
the Earth, too, at 2 m/s2. The apple's acceleration relative to you is 10 m/s2 - 2
m/s2 = 8 m/s2 - the apple seems to fall slower than "normal" free fall.
E. The Elevator Speeds Up - While Going Down - This is exactly the same as Part
D! The direction the elevator moves doesn't matter - only the direction the
elevator accelerates. Since the elevator is accelerating downward in both cases,
the situation inside the elevator is identical!
F. The Elevator Moves Down With Constant Speed - This is exactly the same as
Part C! The direction that the elevator moves doesn't matter - only the elevator's
acceleration. In both cases, the elevator's acceleration is 0 m/s2, so the situation
inside the elevator is the same in both cases.
G. Oh! No!
13. What does the scale read?
a. 500 Newtons
b. 0 Newtons
c. 10 m/s/s (= g)
d. 50 kg (your mass)
e. 10 m/s/s (= g)
f. 500 Newtons
g. 500 Newtons
h. 500 Newtons
i. 0 Newtons
j. 0 Newtons
14. If you let go of the apple, what does it do? Your acceleration and the
apple's acceleration are equal (10 m/s2 downward, so you both fall at the
same rate. Relative to you, the apple stays right where you let go of it - it
appears to float. In other words, the apple appears to be weightless, just as
it would aboard the space shuttle orbiting the earth!
15. What can you do to be weightless? You could jump up in the air, or jump
off a chair. While you are falling, you are weightless! If you jump off a
chair holding an apple, and let go of the apple, the apple will appear to
float right where you left it (relative to you) - it is weightless!