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Physics 6A
Work and Energy examples
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Work and Energy
Energy comes in many forms. We will most often encounter two kinds of energy:
Kinetic Energy – Energy of Motion.
The formula is
Any moving object has kinetic energy.
KE = ½ mv2
Potential Energy – Stored Energy.
There will be several types of potential energy:
* Gravitational – Energy stored by lifting an object above the earth.
more robust formula later, but for now: Ugrav = mgh
We will have a
* Elastic – Energy stored by stretching or compressing a spring.
The formula is Uelastic = ½ kx2
* Electric – Energy stored by charges in an electric field. We will see this next quarter.
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Work and Energy
Work is energy transferred to a moving object when a force acts on it.
To do work, the
force must line up with the motion of the object. Perpendicular forces do no work.
We will have two formulas involving work.
W = Fdcos(θ)
W = ΔKE
Our main concept that ties it all together is Conservation of Energy. This says that
the total energy of a system does not change. We can write down a formula that accounts
for all the forms of energy:
KEi + Ui + WNC = KEf + Uf
This will be the template for most of the problems you will do involving energy.
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Example: A 100kg box (initially at rest) is pushed across a horizontal floor by a
force of 400N. If the coefficients of friction are μk=0.2 and μs=0.4, find the total
work done on the box and the final speed when the box is pushed 10m.
Assume the applied force is horizontal.
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Example: A 100kg box (initially at rest) is pushed across a horizontal floor by a
force of 400N. If the coefficients of friction are μk=0.2 and μs=0.4, find the total
work done on the box and the final speed when the box is pushed 10m.
Assume the applied force is horizontal.
Here is the free-body diagram.
Since the box is moving horizontally, the
only forces that do work are friction and
the 400N push.
Normal
force
friction
400 N
Calculate the force of kinetic friction:
𝑓𝑘 = 𝜇𝑘 𝑚𝑔 = 0.2 100𝑘𝑔
9.8𝑠𝑚2
10m
= 196 𝑁
weight
Work done by each force:
𝑊𝑝𝑢𝑠ℎ = 400𝑁 10𝑚 = 4000 𝐽
𝑊𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = − 196𝑁 10𝑚 = −1960 𝐽
Total work done on box:
𝑊𝑡𝑜𝑡𝑎𝑙 = 2040 𝐽
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Example: A 100kg box on a horizontal floor (initially at rest) is pushed by a force
of 400N, applied downward at an angle of 30°.
If the coefficients of friction are μk=0.2 and μs=0.4, find the total work done on
the box. Does the box move?
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Example: A 100kg box on a horizontal floor (initially at rest) is pushed by a force
of 400N, applied downward at an angle of 30°.
If the coefficients of friction are μk=0.2 and μs=0.4, find the total work done on
the box. Does the box move?
In the last question, we assumed the
box was moving because the problem
told us ho far it moved. This one is
different, and we have to determine
whether or not the box even moves.
To do this, we should find the
maximum friction force and compare to
the forward push – if the push is not
enough to overcome static friction the
box will not move.
Normal
force
friction
30°
400 N
weight
First we will need to break the 400N push into components.
The downward component will increase the normal force.
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Example: A 100kg box on a horizontal floor (initially at rest) is pushed by a force
of 400N, applied downward at an angle of 30°.
If the coefficients of friction are μk=0.2 and μs=0.4, find the total work done on
the box. Does the box move?
Normal
force
First we will need to break the 400N
push into components.
𝑥: 400𝑁 𝑐𝑜𝑠30° = 346𝑁
𝑦: 400𝑁 𝑠𝑖𝑛30° = 200𝑁
To get the max friction we need the
normal force, which will be greater
because of the downward push:
346 N
friction
30°
200 N
400 N
weight
𝐹𝑛𝑜𝑟𝑚𝑎𝑙 = 200𝑁 + 100𝑘𝑔 9.8𝑠𝑚2 = 1180 𝑁
𝑓𝑠,𝑚𝑎𝑥 = 𝜇𝑠 𝐹𝑛𝑜𝑟𝑚𝑎𝑙 = 0.4 1180 = 472 𝑁
Compare this to the forward component of the push (only 346 N). Friction is
strong enough to hold the box in place, so there is no motion.
No work is done on the box.
Note that the actual friction force is only 346N – just enough to holdPrepared
the box.
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Example: A 100kg box is released from rest at the top of a frictionless 10-meter
high ramp that makes an angle of 30° with the horizontal.
Find the final speed of the box when it reaches the bottom of the ramp.
Compare to the impact speed when the box is pushed over the edge and freefalls to the ground instead.
10m
30°
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Example: A 100kg box is released from rest at the top of a frictionless 10-meter
high ramp that makes an angle of 30° with the horizontal.
Find the final speed of the box when it reaches the bottom of the ramp.
Compare to the impact speed when the box is pushed over the edge and freefalls to the ground instead.
We can do this one with conservation of energy.
Here is the basic format:
KEi + Ui + WNC = KEf + Uf
10m
30°
0 + mgh + 0 = ½ mv2 + 0
Solving for v:
𝑣=
2𝑔ℎ =
2(9.8𝑠𝑚2)(10𝑚) = 14𝑚
𝑠
If the box is pushed over the edge, we can use
conservation of energy again, and get the exact
same result. This happens because there was
no friction on the ramp, so the speed at the
bottom is the same in both cases.
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Example: A 100kg box is released from rest at the top of a 10-meter high ramp
that makes an angle of 30° with the horizontal. Assume the coefficients of
friction are μk=0.2 and μs=0.3.
Find the final speed of the box when it reaches the bottom of the ramp.
10m
30°
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Example: A 100kg box is released from rest at the top of a 10-meter high ramp
that makes an angle of 30° with the horizontal. Assume the coefficients of
friction are μk=0.2 and μs=0.3.
Find the final speed of the box when it reaches the bottom of the ramp.
We can do this one with conservation of energy.
Here is the basic format:
KEi + Ui + WNC = KEf + Uf
0 + mgh + Wfriction = ½ mv2 + 0
10m
30°
We need to find the work done by kinetic
friction as the box slides down the ramp. Also,
does it even slide, or is there enough static
friction to hold it in place?
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Example: A 100kg box is released from rest at the top of a 10-meter high ramp
that makes an angle of 30° with the horizontal. Assume the coefficients of
friction are μk=0.2 and μs=0.3.
Find the final speed of the box when it reaches the bottom of the ramp.
Does the box slide at all?
We should draw the free-body diagram to see
what forces are in play. The friction force is
related to the Normal force.
mgsin30
mgcos30
Let’s calculate the max static friction and the
downhill gravity force to see which is bigger.
mg
30°
𝑚𝑔𝑠𝑖𝑛30° = 100𝑘𝑔 9.8𝑠𝑚2 0.5 = 490 𝑁
𝑓𝑠,𝑚𝑎𝑥 = 𝜇𝑠 𝑚𝑔𝑐𝑜𝑠30° = 0.3 100𝑘𝑔 9.8𝑠𝑚2 0.866 = 255 𝑁
Looks like plenty of downhill force to overcome friction. The actual friction
will be kinetic:
𝑓𝑘 = 𝜇𝑘 𝑚𝑔𝑐𝑜𝑠30° = 0.2 100𝑘𝑔 9.8𝑠𝑚2 0.866 = 170 𝑁
As the box slides, kinetic friction will do work against the motion.
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Example: A 100kg box is released from rest at the top of a 10-meter high ramp
that makes an angle of 30° with the horizontal. Assume the coefficients of
friction are μk=0.2 and μs=0.3.
Find the final speed of the box when it reaches the bottom of the ramp.
Work done by kinetic friction:
𝑊𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = −𝑓𝑘 𝑑 = − 170𝑁 20𝑚 = −3400 𝐽
d
10m
Now we can fill in the conservation of energy formula.
0 + mgh + Wfriction = ½ mv2 + 0
100𝑘𝑔 9.8𝑠𝑚2 10𝑚 − 3400𝐽 = 12(100𝑘𝑔)(𝑣 2 )
𝑣 = 11.3
𝑚
𝑠
30°
10𝑚
𝑠𝑖𝑛30° =
𝑑
10𝑚
𝑑=
= 20 𝑚
𝑠𝑖𝑛30°
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1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the
boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s
and slides horizontally without friction.
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1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the
boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s
and slides horizontally without friction.
F=11N
29°
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1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the
boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s
and slides horizontally without friction.
Initially the sled is moving at 0.5 m/s, so its kinetic energy is:


2
K  1 mv 2  1 6.4kg 0.5 m  0.8J
2
2
s
F=11N
29°
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1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the
boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s
and slides horizontally without friction.
Initially the sled is moving at 0.5 m/s, so its kinetic energy is:


2
K  1 mv 2  1 6.4kg 0.5 m  0.8J
2
2
s
Next we can find the work done by the boy’s pull, and add
that to the kinetic energy. Remember – work always equals
the change in the kinetic energy.
F=11N
29°
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1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the
boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s
and slides horizontally without friction.
Initially the sled is moving at 0.5 m/s, so its kinetic energy is:


2
K  1 mv 2  1 6.4kg 0.5 m  0.8J
2
2
s
Next we can find the work done by the boy’s pull, and add
that to the kinetic energy. Remember – work always equals
the change in the kinetic energy.
F=11N
29°
The force is not aligned with the motion, so we
need to use the x-component to get the work.
W  F  cos  d
W  11N  cos29  2m  19.24J
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1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the
boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s
and slides horizontally without friction.
Initially the sled is moving at 0.5 m/s, so its kinetic energy is:


2
K  1 mv 2  1 6.4kg 0.5 m  0.8J
2
2
s
Next we can find the work done by the boy’s pull, and add
that to the kinetic energy. Remember – work always equals
the change in the kinetic energy.
F=11N
29°
The force is not aligned with the motion, so we
need to use the x-component to get the work.
W  F  cos  d
W  11N  cos29  2m  19.24J
Now the total KE is 20.04J. Use this to solve for the new speed:
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1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the
boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s
and slides horizontally without friction.
Initially the sled is moving at 0.5 m/s, so its kinetic energy is:


2
K  1 mv 2  1 6.4kg 0.5 m  0.8J
2
2
s
Next we can find the work done by the boy’s pull, and add
that to the kinetic energy. Remember – work always equals
the change in the kinetic energy.
F=11N
29°
The force is not aligned with the motion, so we
need to use the x-component to get the work.
W  F  cos  d
W  11N  cos29  2m  19.24J
Now the total KE is 20.04J. Use this to solve for the new speed:
K  1 mv 2  20.04J
2
v2 
2(20.04J)
 v  2 .5 m
s
6.4kg
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2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure.
The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its
equilibrium position y=0 and then released, find the ball’s maximum height hmax(measured from the equilibrium
position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all
movement occurs in a straight line up and down along the y-axis.
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2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure.
The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its
equilibrium position y=0 and then released, find the ball’s maximum height hmax(measured from the equilibrium
position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all
movement occurs in a straight line up and down along the y-axis.
[3]
We will use conservation of energy for this one. Notice that the picture already has y=0
defined for us at the top of the tube (this is also the equilibrium position of the spring).
[1]
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2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure.
The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its
equilibrium position y=0 and then released, find the ball’s maximum height hmax(measured from the equilibrium
position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all
movement occurs in a straight line up and down along the y-axis.
[3]
We will use conservation of energy for this one. Notice that the picture already has y=0
defined for us at the top of the tube (this is also the equilibrium position of the spring).
At the beginning (I will call this position [1]) the energy of the system is all potential – the
ball is not moving so K=0. We need to account for the compression of the spring, as
well as the gravitational potential energy of the ball:
[1]
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2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure.
The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its
equilibrium position y=0 and then released, find the ball’s maximum height hmax(measured from the equilibrium
position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all
movement occurs in a straight line up and down along the y-axis.
[3]
We will use conservation of energy for this one. Notice that the picture already has y=0
defined for us at the top of the tube (this is also the equilibrium position of the spring).
At the beginning (I will call this position [1]) the energy of the system is all potential – the
ball is not moving so K=0. We need to account for the compression of the spring, as
well as the gravitational potential energy of the ball:
[1]
E1  1 ky12  mgy1
2


E1  1 667 N  0.25m2  1.5kg  9.8 m2  0.25m
2
m
s 

E1  17.17J
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2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure.
The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its
equilibrium position y=0 and then released, find the ball’s maximum height hmax(measured from the equilibrium
position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all
movement occurs in a straight line up and down along the y-axis.
[3]
We will use conservation of energy for this one. Notice that the picture already has y=0
defined for us at the top of the tube (this is also the equilibrium position of the spring).
At the beginning (I will call this position [1]) the energy of the system is all potential – the
ball is not moving so K=0. We need to account for the compression of the spring, as
well as the gravitational potential energy of the ball:
[1]
E1  1 ky12  mgy1
2


E1  1 667 N  0.25m2  1.5kg  9.8 m2  0.25m
2
m
s 

E1  17.17J
Next we can look at position [3], at the high point. It’s all gravitational potential energy there, since the
ball is not moving (K=0) and the spring is at equilibrium (Uelastic = 0).
E3  mgh max
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2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure.
The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its
equilibrium position y=0 and then released, find the ball’s maximum height hmax(measured from the equilibrium
position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all
movement occurs in a straight line up and down along the y-axis.
[3]
We will use conservation of energy for this one. Notice that the picture already has y=0
defined for us at the top of the tube (this is also the equilibrium position of the spring).
At the beginning (I will call this position [1]) the energy of the system is all potential – the
ball is not moving so K=0. We need to account for the compression of the spring, as
well as the gravitational potential energy of the ball:
[1]
E1  1 ky12  mgy1
2


E1  1 667 N  0.25m2  1.5kg  9.8 m2  0.25m
2
m
s 

E1  17.17J
Next we can look at position [3], at the high point. It’s all gravitational potential energy there, since the
ball is not moving (K=0) and the spring is at equilibrium (Uelastic = 0).
E3  mgh max
Conservation of energy says that the total energy should be the same at both points, so E1 = E3 = 17.17J
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2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure.
The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its
equilibrium position y=0 and then released, find the ball’s maximum height hmax(measured from the equilibrium
position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all
movement occurs in a straight line up and down along the y-axis.
[3]
We will use conservation of energy for this one. Notice that the picture already has y=0
defined for us at the top of the tube (this is also the equilibrium position of the spring).
At the beginning (I will call this position [1]) the energy of the system is all potential – the
ball is not moving so K=0. We need to account for the compression of the spring, as
well as the gravitational potential energy of the ball:
[1]
E1  1 ky12  mgy1
2


E1  1 667 N  0.25m2  1.5kg  9.8 m2  0.25m
2
m
s 

E1  17.17J
Next we can look at position [3], at the high point. It’s all gravitational potential energy there, since the
ball is not moving (K=0) and the spring is at equilibrium (Uelastic = 0).
E3  mgh max
Conservation of energy says that the total energy should be the same at both points, so E1 = E3 = 17.17J
mghmax  17.17J
hmax 
17.17J
 1.17m


m
1.5kg 9.8 2 
s 

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3) A block of mass m1= 2.40kg is connected to a second block of mass m2=1.80kg, as shown on the board. When
the blocks are released from rest, they move through a distance d=0.500m, at which point m2 hits the floor. Give
that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the
blocks just before m2 lands.
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3) A block of mass m1= 2.40kg is connected to a second block of mass m2=1.80kg, as shown on the board. When
the blocks are released from rest, they move through a distance d=0.500m, at which point m2 hits the floor. Give
that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the
blocks just before m2 lands.
We can use conservation of energy. Initially nothing is moving, so K=0
and we only have gravitational potential energy.
Ei  m1gh  m2gd
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3) A block of mass m1= 2.40kg is connected to a second block of mass m2=1.80kg, as shown on the board. When
the blocks are released from rest, they move through a distance d=0.500m, at which point m2 hits the floor. Give
that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the
blocks just before m2 lands.
We can use conservation of energy. Initially nothing is moving, so K=0
and we only have gravitational potential energy.
Ei  m1gh  m2gd
Just before the block lands, both blocks are moving so we will have
kinetic energy, as well as potential energy for block 1 only (block 2 is
now at y=0). Also note that both blocks move at the same speed.
Ef  m1gh  1 m1  m2 v 2
2
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3) A block of mass m1= 2.40kg is connected to a second block of mass m2=1.80kg, as shown on the board. When
the blocks are released from rest, they move through a distance d=0.500m, at which point m2 hits the floor. Give
that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the
blocks just before m2 lands.
We can use conservation of energy. Initially nothing is moving, so K=0
and we only have gravitational potential energy.
Ei  m1gh  m2gd
Just before the block lands, both blocks are moving so we will have
kinetic energy, as well as potential energy for block 1 only (block 2 is
now at y=0). Also note that both blocks move at the same speed.
Ef  m1gh  1 m1  m2 v 2
2
In the absence of friction, we would just set these energies equal. We must account for friction by finding the work
done by friction, and adding it to the initial energy. Note: this work will be negative, so we will end up with less
energy than we started with, as expected (kinetic friction will always take energy away from the system).
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3) A block of mass m1= 2.40kg is connected to a second block of mass m2=1.80kg, as shown on the board. When
the blocks are released from rest, they move through a distance d=0.500m, at which point m2 hits the floor. Give
that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the
blocks just before m2 lands.
We can use conservation of energy. Initially nothing is moving, so K=0
and we only have gravitational potential energy.
Ei  m1gh  m2gd
Just before the block lands, both blocks are moving so we will have
kinetic energy, as well as potential energy for block 1 only (block 2 is
now at y=0). Also note that both blocks move at the same speed.
Ef  m1gh  1 m1  m2 v 2
2
In the absence of friction, we would just set these energies equal. We must account for friction by finding the work
done by friction, and adding it to the initial energy. Note: this work will be negative, so we will end up with less
energy than we started with, as expected (kinetic friction will always take energy away from the system).
Wfric  Ffric  d  cos180
Wfric  k  m1g  d   1
Wfric  k  m1g  d
The friction work is negative because the
force always opposes the motion (that is
why the angle is 180 degrees).
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3) A block of mass m1= 2.40kg is connected to a second block of mass m2=1.80kg, as shown on the board. When
the blocks are released from rest, they move through a distance d=0.500m, at which point m2 hits the floor. Give
that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the
blocks just before m2 lands.
We can use conservation of energy. Initially nothing is moving, so K=0
and we only have gravitational potential energy.
Ei  m1gh  m2gd
Just before the block lands, both blocks are moving so we will have
kinetic energy, as well as potential energy for block 1 only (block 2 is
now at y=0). Also note that both blocks move at the same speed.
Ef  m1gh  1 m1  m2 v 2
2
In the absence of friction, we would just set these energies equal. We must account for friction by finding the work
done by friction, and adding it to the initial energy. Note: this work will be negative, so we will end up with less
energy than we started with, as expected (kinetic friction will always take energy away from the system).
Wfric  Ffric  d  cos180
Wfric  k  m1g  d   1
Wfric  k  m1g  d
The friction work is negative because the
force always opposes the motion (that is
why the angle is 180 degrees).
Finally we just set our final energy equal to the initial energy, plus the (negative) friction work:
Ef  Ei  Wfric
1 m  m v 2  m gd    m g  d
2
2
k
1
2 1
v2 
m2gd  k  m1g  d
 v  1 .3 m
1 m  m 
s
1
2
2
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4) A 95.0 kg diver steps off a diving board and drops into the water 3.00m below. At some depth d below the
water’s surface, the diver comes to rest. If the non-conservative work done on the diver is Wnc= -5120J, what is
the depth ,d?
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4) A 95.0 kg diver steps off a diving board and drops into the water 3.00m below. At some depth d below the
water’s surface, the diver comes to rest. If the non-conservative work done on the diver is Wnc= -5120J, what is
the depth ,d?
We will use conservation of energy. First we must define our coordinate system.
Using y=0 at the lowest point achieved by the diver, we have the following
expressions for the initial and final energy:
Ei  mg(d  h)
Ef  mg(0)  0
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4) A 95.0 kg diver steps off a diving board and drops into the water 3.00m below. At some depth d below the
water’s surface, the diver comes to rest. If the non-conservative work done on the diver is Wnc= -5120J, what is
the depth ,d?
We will use conservation of energy. First we must define our coordinate system.
Using y=0 at the lowest point achieved by the diver, we have the following
expressions for the initial and final energy:
Ei  mg(d  h)
Ef  mg(0)  0
Notice that the kinetic energy is zero in both places because the speed is zero.
The final energy will be the non-conservative work plus the initial energy:
Ef  Ei  Wnc
0  mg(d  h)  5120J
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4) A 95.0 kg diver steps off a diving board and drops into the water 3.00m below. At some depth d below the
water’s surface, the diver comes to rest. If the non-conservative work done on the diver is Wnc= -5120J, what is
the depth ,d?
We will use conservation of energy. First we must define our coordinate system.
Using y=0 at the lowest point achieved by the diver, we have the following
expressions for the initial and final energy:
Ei  mg(d  h)
Ef  mg(0)  0
Notice that the kinetic energy is zero in both places because the speed is zero.
The final energy will be the non-conservative work plus the initial energy:
Ef  Ei  Wnc
0  mg(d  h)  5120J
We can solve this for (d+h), then subtract out the given value for h.
0  mg(d  h)  5120J
( d  h) 
5120J
 5.5m
95kg 9.8 m2

s

d  5.5m  3.0m  2.5m
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