Download Newton`s Laws Review Sheet

Newton’s Laws Review Sheet Answers
Physics
Practice Questions:
1. Give two examples of Newton's first law from class demonstrations and two examples from your own
experience.
Answers will vary for individual students. Some examples from class include:
Dropping the pen in the beaker after removing the ring
Pulling the tablecloth out from under the dishes
Miniature car crash
Skateboarding into a curb
2. You have a hammer with a loose head, which is about to fall off the top. Why is it better to hit the
bottom (butt) of the hammer against the table to tighten it, rather than hitting the top of the hammer
against the table?
This is due to Newton’s First Law. When you hit your hammer against the table, you start the
hammer moving downward. Because of its inertia, it will keep moving downward after that if it
were left on its own. When the handle hits the table, the table provides a force to stop the handle
from moving but the head will keep on moving until something (friction from the handle or the
table) stops it. If you hit the table with the butt end of the hammer, the head of the hammer will
keep moving further onto the handle. If you hit the table with the top end of the hammer, the
head of the hammer will keep moving further off the hammer (though the table will probably
stop it from coming all the way off)
3. What is the difference between an object’s mass and it’s weight? Give examples that illustrate the
difference.
Your weight is a measure of how hard gravity pulls down on you. Your mass is a measure of
how difficult you are to accelerate. Since these two properties are directly related, it can be hard
to tell the difference. One example that illustrates the difference is considering the difference
between pushing a car and lifting a car. When you push a car on flat ground, you feel the effects
of the car’s mass which makes it hard to accelerate (as well as the friction trying to stop the car
from rolling). When you lift a car, you are working against the car’s weight – gravity pulling the
car down. Most people are capable of pushing a car on flat ground, but noone is capable of
lifting a car. Another difference between mass and weight is that an object’s mass is always the
same (providing you don’t add or subtract any matter) while the weight changes depending on
how strong the local gravitational field. On the moon, someone will weight 1/6 of what they do
on Earth, though their mass will be exactly the same
4. Why is it a bad idea to tailgate the car in front of you when you are driving a heavily loaded truck?
A heavily loaded truck has a lot of mass. According to Newton’s 2nd Law, the more mass an
F 

object has the less it will accelerate in response to a force a =
.




This means that a massive truck doesn’t speed up very quickly, but it also means that a massive
truck doesn’t slow down very quickly. A low acceleration when you are stopping means that it
takes a lot of time (and distance) for a massive truck to stop so you want to give yourself lots of
room to be able to stop without hitting the car in front of you. You will notice that good truck
drivers give themselves a lot of space in front of themselves.
m
5. a) A cyclist is pedaling so that a force of 100N pushes her forward while there are 50N of friction
and 50 N of air resistance opposing her movement. What is the net force on the cyclist?
The forward and backwards forces on the bicycle are equal so
they cancel out and leave a net force of 0 N.
b) If she has a mass of 60 kg (including the bicycle), what is her
acceleration?
Fnet 0 N
a = m = 60 kg = 0 m/s/s
c) If she pedals harder so that the force pushing her forward increases to 200N, what is the net force on
her and what is her acceleration?
Fnet = 200 N – 50 N – 50 N = 100 N
Fnet 100 N
a = m = 60 kg = 1.67 m/s/s
6. How much thrust from the engines does a 100,000 kg airplane need to accelerate at 5 m/s/s against
1,000,000 N of friction and air resistance as it starts to take off?
To produce the acceleration, the plane must have enough force acting on it. The net force
required to accelerate the plane is:
Fnet = ma = (100,000 kg)(5 m/s/s) = 500,000 N
Fnet = Fthrust – Ff
(The Ff represents both the friction and the air resistance)
500,000 N = Fthrust – 1,000,000 N
1,500,000 N = Fthrust
7. Jason jumps off a bridge (with a bungee cord attached) and falls until the bungee
cord stretches out and pulls to stop his fall. Jason comes to a stop at the bottom, with
the bungee cord pulling up with 1500 N on his 70 kg mass. How quickly will he
accelerate? Which way?
There are only two forces, the tension in the bungee cord pulling up and gravity pulling
down, so the net force is the difference between those two forces.
Fnet = T – Fg
where Fg = mg = 70 kg x 9.8 m/s/s = 686 N
Fnet = 1500 N – 686 N = 814 N
Then Newton’s 2nd Law gives his acceleration:
Fnet 814 N
a = m = 70 kg = 11.6 m/s/s
8. While visiting family in Iowa, Vanessa finds an old, 25 kg go-cart in their garage. Excited to try
riding, she looks around for a hill, but there aren’t any. To get it moving, she has to push it on the flat
ground. She pushes with 100 N at a 30° angle below horizontal. The coefficient of kinetic friction
against the go-cart is 0.2.
First we look at the situation and then draw a force diagram for the go-cart:
Since Vanessa’s force is at an angle, we need to break it down
into horizontal and vertical components to make it easy to work
with:
a) How much normal force does the ground exert to keep the go-cart
from falling to the center of the earth?
Since the normal force acts in the vertical direction, let’s focus
on the forces in that direction. Since the go-cart isn’t going into
the earth or coming out of the earth, we know that the forces in
the vertical direction have to cancel out:
FN = Fg + 50 N
but Fg = mg = 25 kg x 9.8 m/s/s = 245 N
so,
FN = 245 N + 50 N = 295 N
b) How big is the frictional force acting on the go-cart?
Ff = ·FN = 0.2 · 295 N = 59N
c) What is the net force acting on the go-cart?
Since all of the vertical forces have canceled out, the only forces that contribute to the net force
are in the horizontal direction:
Fnet = 86.6 N – 59 N = 27.6 N
d) How quickly will the go-cart accelerate while she pushes it?
Fnet 27.6 N
a = m = 25 kg = 1.1 m/s/s
9. After working on his tennis ball launcher project at a friend’s house, Greg is hauling his 5 kg project
home in a 15 kg red Radio Flyer wagon. He pulls on the handle with 20 N of force at a 40° angle above
horizontal.
Start by drawing a diagram of the situation and then a force diagram for the wagon:
Find the force of gravity
Fg = mg = (15 kg + 5 kg)(9.8 m/s/s) = 196 N
To simplify things break Greg’s force into components:
a) What is the coefficient of friction between the wagon’s wheels and the ground if that much force pulls
the wagon at constant speed?
Ff
We want to find the coefficient of friction, so we need to rearrange the friction formula:  = F
N
If the wagon is going at constant speed that means that the net force on the wagon is 0 N and all
of the forces cancel out. This will let us find values for all the forces we don’t know.
First look in the vertical direction:
Then in the horizontal direction:
FN + 12.9 N = 196 N
FN = 196 N – 12.9 N = 183.1 N
Ff = 15.3 N
Then find the coefficient of friction:
Ff 15.3 N
 = F = 183.1 N = 0.084
N
9b) What is the coefficient of friction between the wagon’s wheels and the ground if that much force
accelerates the wagon forward at 0.5 m/s/s?
The only thing that is different in part b is that the wagon is moving differently. Everything else
stays the same, so we can use the same free body diagrams that we had up in part a and don’t
have to redraw them! (Yay!)
Ff
Still trying to find the coefficient of friction:  = F
N
If the wagon is accelerating at 0.5 m/s/s that means that the force don’t cancel out. The net force
on the wagon must be Fnet = ma = (20 kg)(0.5 m/s/s) = 10 N
First look in the vertical direction:
Then in the horizontal direction:
Since the wagon is not accelerating in the
vertical direction, the vertical forces must
cancel out. (This looks just like part a)
FN + 12.9 N = 196 N
FN = 196 N – 12.9 N = 183.1 N
Because the wagon is accelerating
forward the forces pulling forward
(horizontal component of Kane’s pull)
must be greater than the forces pulling
backward (friction):
Fnet = 15.3 N – Ff
10 N = 15.3 N – Ff
Ff = 15.3 N – 10 N
Ff = 5.3 N
Then find the coefficient of friction:
Ff
5.3 N
 = F = 183.1 N = 0.029
N
10. Christina rides a cable car up a 30° hill in San Francisco. The 5000 kg cable car is pulled up the
hill by a rope beneath the street. If the coefficient of friction between the cable car’s wheels and the
track is 0.1, how hard does the rope have to pull to keep the cable car at a constant speed?
First we notice that the cable car is moving at a constant speed. That means that all of the forces
cancel out and we can figure out how big any unknown forces are if we know how big the forces
that they cancel out are.
Start with a free body diagram. Since this
problem takes place on a hill, most of the
forces are at angles. To make life easier, we rotate our frame of reference by 30°
(counterclockwise in my picture). This means that all of the forces end up along my new
“vertical” or “horizontal” axes, except gravity, which is 30° away from vertical. I break the
angled gravity force into its “horizontal” and “vertical” components.
We want to find the tension, so we want to look at the forces in the horizontal direction:
We see that the tension is pulling up the
hill against the horizontal/parallel
component of gravity and against friction.
Since the cable car is moving at constant
speed, those forces must cancel out:
T = Fg|| + Ff
Unfortunately, we don’t know how big the
friction force is. Fortunately we can find
out (Ff = ·FN)if we know the normal
force. To find FN, we need to look at the
forces in the vertical direction:
We can see that the normal force has
to cancel out the vertical/perpendicular
component of gravity:
FN = Fg = 42,435 N
That means that the force of friction is:
Ff = ·FN = 0.1 · 42,435 N = 4243.5 N
and:
T = T = Fg|| + Ff = 24,500 N + 4,243.5 N = 28,743.5 N
11. While biking home, Chris passes a spot where the road dips into a
valley between two hills. For kicks he likes to stop pedaling and coast
down the first hill – a 40 m long stretch at a 20° incline - then see if he
can coast all the way up the second hill – a 20 m long stretch at a 30°
incline. He starts at the top of the first hill going 5 m/s. The
coefficient of friction against the bicycle is 0.1. Together, Chris and
his bike mass 70 kg.
The first 3 questions (a,b,c) involve Chris speeding up as he rolls down the first hill. For this
section, we need to look at the forces acting on the downhill section. The last 3 questions (d,e,f)
pick up at the bottom of the two hills, as Chris is about to start up the second hill. On the way up
the second hill, the forces will act to slow Chris down. This means that we need to draw
diagrams for the two sections separately.
a) What is the net force acting on Chris as he rolls down the first hill?
Clearly, gravity pulls down: Fg = mg = 70 kg · 9.8 m/s/s = 686 N.
There is also the normal force pointing directly off the hillside
and friction against the motion (up the hill)
Because this problem involves a hill, we are going to rotate our
frame of reference to match up with the hillside. Now the
friction and normal force are directly “horizontal” and “vertical”
and only gravity is at an angle. We must break the gravitational
force into components.
Fg = 686 N·cos(20°) = 644.6 N
Fg|| = 686 N·sin(20°) = 234.6 N
The perpendicular component of gravity opposes the normal force. Because there is no
acceleration in the “vertical” direction those two forces must cancel out:
FN = Fg = 644.6 N
With the “vertical” forces canceling out, all that is left is the “horizontal” forces. The
“horizontal” component of gravity is what speeds him up. It is resisted by friction:
Ff = ·FN = 0.1 · 644.6 N = 64.5 N
The net force is then what is left when the friction is canceled out of the parallel component of
gravity:
Fnet = Fg|| – Ff = 234.6 N – 64.5 N = 170.1 N down the hill
b) What is Chris’s acceleration as he rolls down the first hill?
Fnet 170.1 N
a = m = 70 kg = 2.43 m/s/s down the hill
c) How fast is Chris going when he reaches the bottom of the hill?
We set positive direction down the hill because everything goes that way.
Then vi = +5 m/s; d = +40 m; a = +2.43 m/s/s
vf2 = vi2 + 2a·d
vf2 = (5 m/s)2 + 2·2.43 m/s/s·40 m
vf2 = 219 m2/s2
vf = 14.8 m/s
d) What is the net force acting on Chris as he rolls up the second hill?
The big difference between going up a hill and going down is
that when you go up a hill, the friction against your motion is
pointing in the same direction as the parallel component of
gravity instead of against it.
First we reorient out frame of reference and break the force of
gravity into components. Since Chris’s mass is still the same
Fg = 686 N·cos(30°) = 594.1 N
Fg|| = 686 N·sin(30°) = 343 N
The perpendicular component of gravity opposes the normal force. Because there is no
acceleration in the “vertical” direction those two forces must cancel out:
FN = Fg = 594.1 N
With the “vertical” forces canceling out, all that is left is the “horizontal” forces. The
“horizontal” component of gravity works with friction to slow him down:
Ff = ·FN = 0.1 · 594.1 N = 59.4 N
The net force is then the total of friction and the parallel component of gravity:
Fnet = Fg|| + Ff = 343 N + 59.4 N = 402.4 N down the hill
e) What is Chris’s acceleration as rolls up the second hill?
Fnet 402.4 N
a = m = 70 kg = 5.75 m/s/s down the hill
f) How far up the second hill will Chris make it without pedaling?
First we set up our frame of reference so that positive is up the hill – the direction he starts out
moving. See the picture below and to the right. vi = 14.8 m/s; a = -5.75 m/s/s
Now that we are ready, there are two ways to approach the question:
The first is ask whether Chris makes it to the top or not. If he does make it to the top, then his
final velocity when he goes 20 m will not be zero.
We pick a direction and then
vf2 = vi2 + 2a·d
vf2 = (14.8 m/s)2 + 2·(-5.75 m/s/s)·20 m
vf2 = -11.0 m2/s2 and the square root of a negative number is
not real. This means that he does not make it to the top of the
hill.
The second approach is to ask how far he goes before coming
to a stop: when vf = 0 m/s.
vf2 = vi2 + 2a·d
02 =(14.8 m/s)2 + 2(-5.75 m/s/s)d
0 = 219.04 m2/s2 –11.5 m/s/s·d
11.5 m/s/s·d = 219.04 m2/s2
d = 19 m only 1 m shy of the top!
12. Why is it always a good idea to tie your canoe to the dock before you try to step from the canoe onto
the dock? What happens to the canoe if it’s not tied up? Why?
When you step out of the canoe, you do so by pushing backward on the canoe. Because of
Newton’s 3rd Law, it reacts by pushing you forward and out of the canoe, up onto the dock.
When you push backward on the canoe, the canoe will accelerate backward if there is no other
force to hold it in place. If it is tied to the dock, the rope can exert a force to hold the canoe in
place.
Having the canoe tied up, gives you another advantage. Because a canoe is usually significantly
less massive than the person who is trying to step out, it will often accelerate away under very
little force. If you only exert a small force on the canoe, it will only exert a small force on you,
which may not accelerate you quickly enough to get to the dock before the canoe slips out from
under you (splash). Having the rope tied to the dock supports the canoe (by canceling out your
force on the canoe) so that it can exert a greater force on you without producing a net force on
the canoe and accelerating the canoe away from the dock. This also accelerates you more
rapidly to get you safely on the dock quickly.
13. When a pitcher throws a baseball, the pitcher exerts a significant force on the baseball. What is the
reaction to that force? Why don’t we normally see that reaction?
The reaction to the pitcher’s force is that the ball pushes backwards on the pitcher with the same
amount of force that the pitcher put on the ball. There are two reasons why we don’t notice that
the ball is pushing on the pitcher:
1. The pitcher is planted on the ground, and the friction between the pitcher’s feet and the
ground cancels out the force of the ball. This means that the ball doesn’t make the pitcher
accelerate backwards.
2. The pitcher is more massive than the ball. Even though the ball doesn’t make the pitcher
accelerate backwards, it still acts to slow down the pitcher’s movement forward. However, this
is essentially unnoticeable, because the pitcher has a large mass (compared to the ball) but is
affected by a force the same size that the ball felt. A low mass object accelerates a lot in
response to a force, while a high mass object accelerates a little in response to the same force.
The ball speeds up a lot. The pitcher slows down very little.
14. Why is it harder to run through water than through air?
When you run through water or air, you have to push on the air and water and make it accelerate
to get it out of your way. Because of Newton’s 3rd Law, the air and water pushes back on you
and tries to stop you from moving.
Because water is a lot more massive than the same volume of air, you have to push a lot harder
to accelerate a particular amount of water out of your way than you would to get the same
amount of air out of your way. This is one of the points in Newton’s 2nd Law.
According to Newton’s 3rd Law, the harder you push on something, the harder it pushes back on
you. This means that as you push harder to get the water to accelerate out of your way, it pushes
back on you harder than the air would, which makes it harder for you to run.
15. Kevin gets hired as an astronaut and is sent on a mission to fix the Hubble Space Telescope. While
he is free-walking in space to work on the Telescope, he discovers that he is drifting slowly (but surely)
away from the telescope and the space shuttle, with nothing but a wrench. How can he use his
knowledge of physics to get back to the ship?
Obviously, Kevin will apply his knowledge of Newton’s 3rd Law. He knows that when he
pushes on the wrench, the wrench pushes back on him. This means all he has to do is push the
wrench away from the space shuttle and the wrench will push him back toward the space shuttle.
The real problem is being able to apply enough force to a low mass object like a wrench, which
will quickly accelerate away from him as he pushes it. Because his mass is relatively large, he
needs a relatively large force to produce any appreciable change in his speed.
If he can’t get enough acceleration from the wrench, he might try cracking the seal on his space
suit on the side away from the space shuttle. As the space suit pushes the air inside out through
the crack away from the space shuttle, the will push the spacesuit (and Kevin) back toward the
space shuttle. The air actually pushes surprisingly hard (because it gets accelerated quickly out
of the hole so there is a pretty large force accelerating him toward the space shuttle. The real
trick would be not losing too much air so that you could still breathe on the trip back.
16. Justin (70 kg) is exiled to a small – 100,000,000 kg – asteroid, which has no significant gravity, for
his misbehavior. There’s not much to do on an asteroid that small, so he gets terribly bored.
One day he stamps his foot in frustration and accelerates from rest to going up at 1 m/s in
0.2 seconds.
a) How hard does the asteroid push on him when he stamps his foot? (How quickly is he
accelerating?)
We can calculate how hard the asteroid pushes on Justin using Newton’s 2nd
Law (F = ma) if we know how quickly he is accelerating.
vf – vi 1 m/s – 0 m/s
a= t =
= 5 m/s/s
0.2 s
Since it is Justin who is being accelerated, we use his mass – 70 kg – in the calculation:
Fnet = ma = (70 kg)(5 m/s/s) = 350 N
b) How hard does Justin push on the asteroid? Which way?
By Newton’s 3rd Law we know that Justin pushes on the asteroid just as
hard as the asteroid pushes on Justin, only in the opposite direction. So the
force applied to the asteroid is 350 N “downward” in my picture.
c) How fast is the asteroid moving after Justin’s stamp?
Since the asteroid pushed Justin for 0.2 seconds, Justin pushed the asteroid
for 0.2 seconds. The we can use our motion equations to find the asteroid’s
final velocity, if we just know how quickly it accelerated. That calls for Newton’s 2nd Law.
Since we are talking about the asteroid’s motion here, we use the asteroid’s mass:
Fnet
350 N
a = m = 100000000 kg = 0.00000350 m/s/s
We then plug that information into the second equation of motion:
vf = vi + at = 0 m/s + (0.00000350 m/s/s)0.2 s = 0.000007 m/s, which is not terribly fast.
17. One of the duties of a good father is to take his children out roller skating for
their entertainment. A 90 kg father is skating at 2 m/s behind his 25 kg daughter who
is going the same speed. He gives her a quick (0.5 s) push and winds up sitting still
afterward. The coefficient of friction against the roller skates’ motion is 0.1.
a) How hard does the daughter push the father to bring him to a stop?
For the father to come to a stop, he has to accelerate:
a=
vf – vi 0 m/s – 2 m/s
= 4 m/s/s
t =
0.5 s
This acceleration is produced by the push from his daughter and from friction working together.
Together they have to equal:
Fnet = ma = 90 kg · 4 m/s/s = 360 N
Now since friction is:
Ff = ·FN = 0.1 · 882 = 88.2 N
(On flat ground the normal force balances gravity:
FN = Fg = mg = 90 kg · 9.8 m/s/s = 882 N)
That means that to get enough force the daughter has to push:
Fnet = Fdaughter + Ff
360 N = Fdaughter + 88.2 N
Fdaughter = 271.8 N ←
b) How hard does the father push the daughter?
Because of Newton’s 3rd Law, the father pushes the daughter with exactly the same force she
applied to him, only pointing the opposite direction: Ffather = 271.8 N→
c) How fast is the daughter going after the push?
The daughter will accelerate because of the force of the father’s push, though friction will resist
that acceleration.
The amount of friction acting on the daughter is
Ff = ·FN = 0.1 · 245 N = 24.5 N
(On flat ground the normal force balances gravity:
FN = Fg = mg = 25 kg · 9.8 m/s/s = 245 N)
The net force acting on the daughter is then:
Fnet = Ffather – Ff = 271.8 N – 24.5 N = 247.3 N
This force will produce an acceleration of:
Fnet 247.3 N
a = m = 25 kg = 9.89 m/s/s
At that rate, she will speed up for 0.5 s, until her final velocity is:
vf = vi +at = 2 m/s + 9.89 m/s/s · 0.5 s = 6.945 m/s
18. A commuter train consists of an engine and two passenger cars. After loading the passengers at a
station, car A has a mass of 14,000 kg and car B has a mass of 13,000 kg. The train begins to speed up
and is going 30 m/s after 5 seconds. The coefficient of kinetic friction against the cars’ motion is 0.15.
a) What is the train’s acceleration?
a=
vf – vi 30 m/s – 0 m/s
= 6 m/s/s
t =
5s
This means that each car and the engine are all accelerating at 6 m/s/s together.
b) How hard does Car A have to pull on Car B to make it accelerate?
For the car to accelerate, there has to be an unbalanced for acting on it:
Fnet = ma = 13,000 kg · 6 m/s/s = 78,000 N
This force is produced by the force applied by the car in
front of it – Car A – overcoming the friction acting on
it.
To calculate the friction, we need to know the normal
force, which is acting to cancel out gravity in this
situation:
FN = Fg = mg = 13,000 kg · 9.8 m/s/s = 127,400 N
Ff = ·FN = 0.15 · 127,400 N = 19,110 N
Then:
Fnet = FCar A – Ff
78,000 N = FCar A – 19,110 N
FCar A = 97,110 N
c) How hard does the engine have to pull on Car A to make it accelerate?
The lead car is pulled forward by the engine against the resistance of friction and the second car.
While Car A is pulling Car B forward, Car B pulls Car A backward just as hard, with 97,110 N
to the right.
To calculate the friction, we need to know the normal force, which is acting to cancel out gravity
in this situation:
FN = Fg = mg = 14,000 kg · 9.8 m/s/s = 137,200 N
Ff = ·FN = 0.15 · 137,200 N = 20,580 N
The net force to make Car A accelerate is:
Fnet = ma = 14,000 kg · 6 m/s/s = 84,000 N
Fnet = Fengine - FCar B – Ff
84,000 N = Fengine – 97,110 N – 20,580 N
Fengine = 201,690 N
19. Matt is out climbing on Mount Hood when he
comes to the edge of 20 m high cliff. He ties his rope to
a rock and lowers his 70 kg body over the edge of the
cliff. Unfortunately for him the 110 kg rock is just
resting on the ground, with a coefficient of kinetic
friction of 0.6 between it and the ground.
a) How quickly will Matt and the rock accelerate?
Since Matt and rock are tied together, they will
move together, both accelerating at the same
rate. This means that we can treat them as one
object and just add up all the forces acting on
them. The force of gravity on Matt speeds them up while the friction on the rock slows them
down. It is a little bit odd that the two forces that are opposing each other are actually
perpendicular – gravity pulls down while the friction on the rock points to the right, but that
happens because a rope can pull around the corner and change the direction that it exerts its
tension on either side of the corner. A little weird and yet we use that fact all the time.
Therefore the net force on the combined system of Matt and the rock is:
Fnet = Fg-Matt – Ff-rock
And we fill in the blanks:
Fg-Matt = mg = 70 kg ·9.8 m/s/s = 686 N
The force of friction on the rock depends on the normal force on the rock – which is exactly
countering the gravity pulling down on the rock:
FN = Fg-rock = mg = 110 kg · 9.8 m/s/s = 1078 N
Ff = ·FN = 0.6 · 1078 N = 646.8 N
Then:
Fnet = Fg-Matt – Ff-rock
Fnet = 686 N – 646.8 N = 39.2 N
This net force will make them accelerate according to Newton’s 2nd Law, but since we are using
both objects combined together, we must use the total mass: m = 70 kg + 110 kg = 180 kg
Fnet 39.2 N
a = m = 180 kg = 0.218 m/s/s
b) Assuming that the rope is long enough and that the rock doesn’t fall over the edge of the cliff and
squash Matt, how quickly will Matt be moving when he reaches the ground?
There are 20 m to fall down, accelerating downward at 0.22 m/s/s, starting from rest. Since
everything moves down, we will call down the positive direction and all our numbers are
positive:
vf2 = vi2 + 2a·d
vf2 = (0 m/s)2 + 2·(0.218 m/s/s)·20 m
vf2 = 8.72 m2/s2
vf2 = 2.95 m/s
Nice and slow. Not a bad way to get down, as long as the rock doesn’t fall on you.
c) How much tension is in Matt’s rope as he moves toward the ground? (How hard does it pull on
Matt? How hard does it pull on the rock?)
Though it asks how hard the rope pulls on Matt and how hard the rope pulls on the rock, it is
really one question, since the rope is just the way that Matt and rock pull on each other and they
should pull on each other equally hard according to Newton’s 3rd Law. Let’s prove this by
considering each object separately:
Tension on the rock:
The rock is accelerating at 0.22 m/s/s, meaning that the unbalanced force acting on it is:
Fnet = ma = 110 kg · 0.218 m/s/s = 24.0 N
This unbalanced force is produced when friction cancels out
most of the tension in the string: (FN and Fg cancel and are not
important)
Fnet = T – Ff
24.0 N = T – 646.8 N
T = 670.8 N
Tension on Matt:
Matt is accelerating at 0.22 m/s/s, meaning that the amount of unbalanced force acting on him is:
Fnet = ma = 70 kg · 0.218 m/s/s = 15.3 N
This unbalanced force is produced when the rope’s tension cancels out
most of the gravity on Matt:
Fnet = Fg – T
15.3 N = 686 N – T
T = 670.7 N
These two results – 670.8 and 670.7 – are essentially the same and would be exactly the same if
we didn’t round our acceleration. This demonstrates Newton’s 3rd Law at work – the rock pulls
back on Matt just as hard as Matt pulls on the rock.