Mechanics - College Physics
Card 0 of 20
A toy car is set up on a frictionless track containing a downward sloping ramp and a vertically oriented loop. Assume the ramp is 
tall. The car starts at the top of the ramp at rest.
What additional piece of information is necessary to calculate the maximum height of the loop if the car is to complete the loop and continue out the other side?
A toy car is set up on a frictionless track containing a downward sloping ramp and a vertically oriented loop. Assume the ramp is
What additional piece of information is necessary to calculate the maximum height of the loop if the car is to complete the loop and continue out the other side?
This is an example of conservation of energy. The car starts at the top of the ramp, at height 
. It has no velocity at this time since it is starting from a rest. Therefore its total energy is 
where 
is the mass of the car and 
is the value of gravitational acceleration.
At the bottom of the loop, all of the potential energy will have been converted into kinetic energy.
As the car traverses the loop and rises above the ground, kinetic energy will be converted back into potential energy. The shape of the loop does not matter in this case -- only the vertical distance between the ground and the car.
In the tallest possible loop, all kinetic energy at the bottom is converted to potential energy at the top. This is the maximum height the car can reach -- there is no additional energy left to continue climbing a taller loop. Therefore, the potential energy at the top of the tallest loop we can build is equal to the kinetic energy at the bottom of the loop. But we have already noted that the kinetic energy at the bottom of the loop is equal to the potential energy at the top of the ramp.
Therefore, we set 
. We see that 
and 
cancel, and we are left with 
. In other words, the tallest loop you can build is equal to the height of whatever ramp you select. In this example, the tallest loop we can build is 
. We do not need to know the specific values of 
or 
.
This is an example of conservation of energy. The car starts at the top of the ramp, at height
At the bottom of the loop, all of the potential energy will have been converted into kinetic energy.
As the car traverses the loop and rises above the ground, kinetic energy will be converted back into potential energy. The shape of the loop does not matter in this case -- only the vertical distance between the ground and the car.
In the tallest possible loop, all kinetic energy at the bottom is converted to potential energy at the top. This is the maximum height the car can reach -- there is no additional energy left to continue climbing a taller loop. Therefore, the potential energy at the top of the tallest loop we can build is equal to the kinetic energy at the bottom of the loop. But we have already noted that the kinetic energy at the bottom of the loop is equal to the potential energy at the top of the ramp.
Therefore, we set
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An elevator is designed to hold 
of cargo. The designers want the elevator to be able to go from the ground floor to the top of a 
tall building in 
. What is the minimum amount of power that must be delivered to the motor at the top of the shaft? Assume no friction and that the elevator itself has a negligible weight.
An elevator is designed to hold
Power is the rate of energy transfer. To raise a 
object 
, a total of 
or (
is required. To find the power in Watts (
), we divide the total energy required by the time over which the energy must be transferred:
Power is the rate of energy transfer. To raise a
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How far can a person jump while running at 
and a vertical velocity of 
?
How far can a person jump while running at
We know that:
and we are looking for the maximum height (vertical displacement) this person can obtain, so we aren't concerned with 
.
We can apply the conservation of energy:
Masses cancel, so
Solve for 
:
(rounded to simplify our calculations)
so let's plug in what we know
. This is our final answer.
We know that:
and we are looking for the maximum height (vertical displacement) this person can obtain, so we aren't concerned with
We can apply the conservation of energy:
Masses cancel, so
Solve for
so let's plug in what we know
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If a 
object has a kinetic energy of 
right after it is launched in the air, and it has 
KE at its max height, what is its max height?
If a
Let's first write down the information we are given:
In order to solve this problem we must apply the conservation of energy, which states 
since no friction.
This means that as the project reaches its max height energy is converted from Kinetic Energy (energy of motion) to potential gravitational energy (based off of height).
We can subtract 
from 
to get the 
at its max height
=
so we can solve for the height
where 
therefore 
Let's first write down the information we are given:
In order to solve this problem we must apply the conservation of energy, which states
This means that as the project reaches its max height energy is converted from Kinetic Energy (energy of motion) to potential gravitational energy (based off of height).
We can subtract
so we can solve for the height
where
therefore
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If an object has a kinetic energy of 
right after it is launched in the air, and it has 
KE at its max height of 
, what is the object's mass?
If an object has a kinetic energy of
Let's first write down the information we are given:
In order to solve this problem we must apply the conservation of energy, which states since no friction.
This means that as the project reaches its max height energy is converted from Kinetic energy (energy of motion) to potential gravitational energy (based off of height).
We can subtract from to get the at its max height
=
so we can solve for the mass
where 
therefore 
Let's first write down the information we are given:
In order to solve this problem we must apply the conservation of energy, which states
This means that as the project reaches its max height energy is converted from Kinetic energy (energy of motion) to potential gravitational energy (based off of height).
We can subtract
so we can solve for the mass
where
therefore
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Fluid flows into a pipe with a diameter of 
at a rate of 
. If the other end of the pipe has a cross sectional area of 
, what is the speed of the fluid as it exits the pipe?
Fluid flows into a pipe with a diameter of
This question tests the concept of the continuity equation which stipulates that at steady state the volume flowing into one end of a pipe must be exactly equal to the volume flowing out of the other end. That is, 
. Note that in this case, we are given the inlet diameter and not cross-sectional area. Thus, we must also find the inlet's cross-sectional area by using the formula for the area of a circle.
This question tests the concept of the continuity equation which stipulates that at steady state the volume flowing into one end of a pipe must be exactly equal to the volume flowing out of the other end. That is,
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If the radius of a pipe with flowing liquid is decreased by a factor of 
, how does the velocity of the fluid need to change in order to maintain a constant flow rate?
If the radius of a pipe with flowing liquid is decreased by a factor of
To see how a change of radius affects the velocity of a moving fluid, we'll need to take the continuity equation into account.
Even though the radius is not shown in the above expression, remember that area is related to radius.
Or put another way:
When substituted into the continuity equation:
As the above proportionality shows, if the radius were to decrease by a factor of 
, this would cause the flow rate to decrease by a factor of 
. To counteract this drop in flow rate, the velocity term would need to increase by a factor of 
.
To see how a change of radius affects the velocity of a moving fluid, we'll need to take the continuity equation into account.
Even though the radius is not shown in the above expression, remember that area is related to radius.
Or put another way:
When substituted into the continuity equation:
As the above proportionality shows, if the radius were to decrease by a factor of
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The reason why a coconut floats in water is because __________.
The reason why a coconut floats in water is because __________.
The equation for the buoyant force is 
. The gravitational force is constant. In order for the coconut to float, the buoyant force must be greater than the gravitational force. If the gravitational force is greater than the buoyant force, then the object will sink.
The equation for the buoyant force is
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A baseball has a mass of 
, but it weighs 
when completely submerged in water. What is its volume assuming that the density of water is 
?
A baseball has a mass of
We are given the mass of the baseball outside of the water. Using the weight equation 
with the gravitational constant being 
and the mass being 
, the weight of the baseball outside of the water is 4.905 N. (Be careful and convert the mass of the baseball from grams to kilograms since we are using SI units).
The buoyancy force is the difference of the weight of the baseball when it is in the air and when it is in the water. So subtract the two differences:
Now we use the buoyancy equation:
where 
is the buoyancy force, 
is the density of water, 
is the volume of the baseball, and 
is the gravitational constant. Plug in the known variables and solve for the volume.
and we get 
We are given the mass of the baseball outside of the water. Using the weight equation
The buoyancy force is the difference of the weight of the baseball when it is in the air and when it is in the water. So subtract the two differences:
Now we use the buoyancy equation:
and we get
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Suppose a 
box is being dragged across the floor due to a rope being pulled on at a 
angle from the side, as shown in the picture below.
If the tension in the rope is 
and the box accelerates to the right at 
, what is the coefficient of kinetic friction?
Suppose a
If the tension in the rope is
To solve this problem, we first need to take into account the forces acting in the vertical direction separately from the forces acting in the horizontal direction.
First, let's start with the vertical direction. Here, the only force force acting downward is the weight of the box. There are two forces acting upwards on the box; one is the normal force and the other is the vertical component of the tension in the rope that is pulling the box. Because the box is not moving in the vertical direction, there is no net force. Thus, the sum of these forces is equal to zero.
Next, let's take a look at the forces acting horizontally on the box. Acting to the right of the box is the horizontal component of the tension in the rope. Acting to the left is the frictional force. Because the box is moving to the right, it must have experienced a net force in this direction. Thus, the sum of these horizontally acting forces will equal a net force.
Now, we can take the expression we obtained for the normal force and substitute it into the expression we obtained for the horizontally acting forces.
Now that we've found an expression for the coefficient of kinetic friction, all we need to do is plug in the values given in the question stem to arrive at the answer.
To solve this problem, we first need to take into account the forces acting in the vertical direction separately from the forces acting in the horizontal direction.
First, let's start with the vertical direction. Here, the only force force acting downward is the weight of the box. There are two forces acting upwards on the box; one is the normal force and the other is the vertical component of the tension in the rope that is pulling the box. Because the box is not moving in the vertical direction, there is no net force. Thus, the sum of these forces is equal to zero.
Next, let's take a look at the forces acting horizontally on the box. Acting to the right of the box is the horizontal component of the tension in the rope. Acting to the left is the frictional force. Because the box is moving to the right, it must have experienced a net force in this direction. Thus, the sum of these horizontally acting forces will equal a net force.
Now, we can take the expression we obtained for the normal force and substitute it into the expression we obtained for the horizontally acting forces.
Now that we've found an expression for the coefficient of kinetic friction, all we need to do is plug in the values given in the question stem to arrive at the answer.
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A baseball weighing 
is dropped from a second story window which is 
high. What is the gravitational potential energy?
A baseball weighing
Gravitational potential energy is given by the equation:
We are given all the information needed to solve for the potential energy.
Plug in known values and solve.
Recall that the units for energy are Joules. Newtons is the unit for force.
Gravitational potential energy is given by the equation:
We are given all the information needed to solve for the potential energy.
Plug in known values and solve.
Recall that the units for energy are Joules. Newtons is the unit for force.
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A 
bowling ball is released from 
in the air. What is it's gravitation potential energy upon release?
A
The equation for gravitational potential energy is:
We are given all the information needed to answer the question.
Plug in known values and solve.
The equation for gravitational potential energy is:
We are given all the information needed to answer the question.
Plug in known values and solve.
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A 
bouncy ball is dropped from 
. When it bounces back up is reaches a height of 
. How much energy was loss?
A
The formula for gravitational potential energy is:
To find the potential energy lost, we need to find the potential energy of the ball at two heights, then find the difference.
Note that the energy was not actually lost; rather, it was converted to kinetic energy.
The formula for gravitational potential energy is:
To find the potential energy lost, we need to find the potential energy of the ball at two heights, then find the difference.
Note that the energy was not actually lost; rather, it was converted to kinetic energy.
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A pendulum is made up of a small 
mass that hangs on the end of a 
long string of negligible mass. The pendulum is displaced by 
and allowed to undergo harmonic motion. What is the angular frequency of the resulting motion?
A pendulum is made up of a small
The angular frequency of a simple pendulum is 
, where 
is the length of the pendulum.
The angular frequency of a simple pendulum is
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For a simple harmonic motion governed by Hooke's Law, 
, if 
is the period then the quantity 
is equivalent to which of the following?
For a simple harmonic motion governed by Hooke's Law,
We know that T is the period. The equation for T is 
for harmonic motion.
Solve for 
by dividing the equation by 
on both sides. The result is 
, which is the answer.
We know that T is the period. The equation for T is
Solve for
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If the mass of a simple pendulum is quadrupled, then its period __________.
If the mass of a simple pendulum is quadrupled, then its period __________.
We know that the equation for the period of a simple pendulum is 
. This equation does not depend on mass. It is only affected by the length of the pendulum (L) and the gravitational constant (g). Therefore, adding mass to the pendulum will not effect the period, so the period remains the same.
We know that the equation for the period of a simple pendulum is
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A violin string 
long has a linear density of 
. What is the string tension if the second harmonic has a frequency of 
?
A violin string
Since we are solving for string tension, we need to use the frequency equation with the tension variable in it. That equation is 
where 
is the frequency, 
is the number of the harmonic, 
is the length of the string, 
is the linear density of the string, and 
is the tension of the string.
We are given:
Next we must convert the length of 
to meters which is 
and the mass density of 
to 
. Then we plug in our known values into the equation and solve for the string tension. The result is 
.
Since we are solving for string tension, we need to use the frequency equation with the tension variable in it. That equation is
We are given:
Next we must convert the length of
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A pendulum on earth has a period of 
. What is it's period on Mars with it's gravity is 
A pendulum on earth has a period of
First, we need to find the length of the pendulum. Begin with the equation for finding the period of a pendulum:
solve for 
to get:
Now we can plug in our given values:
Since we have the length of the pendulum determined, we can now find the period of the pendulum on Mars:
First, we need to find the length of the pendulum. Begin with the equation for finding the period of a pendulum:
solve for
Now we can plug in our given values:
Since we have the length of the pendulum determined, we can now find the period of the pendulum on Mars:
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Which of the following is not an example of simple harmonic motion?
Which of the following is not an example of simple harmonic motion?
For this question, we need to recall what simple harmonic motion is. Remember that it is a periodic motion where the restoring force depends on the displacement of the object undergoing these motions. So to answer this question, we need to keep this idea in mind and see which example doesn't match up.
A mass on a pendulum moving back and forth is clearly an example of simple harmonic motion. As the mass moves further from the center in either direction, it experiences a greater and greater force in the opposite direction.
A child swinging on a swing set is another correct example. This situation is analogous to the mass on a pendulum swinging back and forth.
A vibrating guitar string is yet another example of simple harmonic motion. After it is plucked, the string oscillates back and forth.
Finally, a book falling to the ground does not represent harmonic motion. Once the book is released from rest, we intuitively know that it will fall to the ground and will then stay there; in no way is there any periodic motion.
For this question, we need to recall what simple harmonic motion is. Remember that it is a periodic motion where the restoring force depends on the displacement of the object undergoing these motions. So to answer this question, we need to keep this idea in mind and see which example doesn't match up.
A mass on a pendulum moving back and forth is clearly an example of simple harmonic motion. As the mass moves further from the center in either direction, it experiences a greater and greater force in the opposite direction.
A child swinging on a swing set is another correct example. This situation is analogous to the mass on a pendulum swinging back and forth.
A vibrating guitar string is yet another example of simple harmonic motion. After it is plucked, the string oscillates back and forth.
Finally, a book falling to the ground does not represent harmonic motion. Once the book is released from rest, we intuitively know that it will fall to the ground and will then stay there; in no way is there any periodic motion.
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A car traveling at 
has a kinetic energy 
. If the car accelerates to 
, what will the new kinetic energy be?
A car traveling at
Kinetic energy is given by
. We will begin by calculating the car's initial kinetic energy, in terms of the unknown mass of the car 
:
.
Next, we will calculate the final energy of the car, also in terms of the unknown mass of the car:
.
To find the ratio of the final to initial kinetic energy, we divide 
by 
. We see that this reduces to 
with the both the mass 
and 
terms cancelling. 
. Thus, the new kinetic energy is 
.
Kinetic energy is given by
Next, we will calculate the final energy of the car, also in terms of the unknown mass of the car:
To find the ratio of the final to initial kinetic energy, we divide
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