Forces - Physics
Card 0 of 80
An hourglass is placed on a scale with all its sand in the upper chamber. A short time later, the sand begins to fall into the lower chamber. Which of the following best describes the reading on the scale as a function of time before any sand has accumulated in the bottom chamber?
An hourglass is placed on a scale with all its sand in the upper chamber. A short time later, the sand begins to fall into the lower chamber. Which of the following best describes the reading on the scale as a function of time before any sand has accumulated in the bottom chamber?
Initially, when all the sand is in the upper chamber, the reading on the scale is constant and corresponds to the weight of the hourglass and the sand within. As some sand falls, there is no normal force on it so the scale can not register its weight. The fraction of sand that is falling is small, so there is only a small decrease in the apparent weight of the hourglass.
Initially, when all the sand is in the upper chamber, the reading on the scale is constant and corresponds to the weight of the hourglass and the sand within. As some sand falls, there is no normal force on it so the scale can not register its weight. The fraction of sand that is falling is small, so there is only a small decrease in the apparent weight of the hourglass.
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A juggler throws a ball straight up into the air, which exerts air resistance on the ball. Which of the following best describes the speed of the ball after it falls and just reaches the juggler's hand?
A juggler throws a ball straight up into the air, which exerts air resistance on the ball. Which of the following best describes the speed of the ball after it falls and just reaches the juggler's hand?
Air resistance is like a friction force that takes energy away from the ball. Because the ball has energy taken from it on the way up and the way down, there is net negative work done on the ball. The energy of the ball is not conserved and it slows down slightly compared to when it was released.
Air resistance is like a friction force that takes energy away from the ball. Because the ball has energy taken from it on the way up and the way down, there is net negative work done on the ball. The energy of the ball is not conserved and it slows down slightly compared to when it was released.
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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 
cinder block sitting on a frictionless surface is attached to a rope. If the rope is pulled parallel to the surface such that the block accelerates at a rate of 
, what is the minimum amount of tension the rope must be able to support without breaking?
A
This is a simple case of 
. In this instance, 
is 
and 
is 
. Substituting, we find 
.
This is a simple case of
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A 
cinder block sitting on the ground is attached to a rope. If the rope is pulled such that the block accelerates directly upwards at a rate of 
, what is the minimum amount of tension the rope must be able to support without breaking?
A
In this case, the rope must be able to support both the force of gravity and the force required to accelerate the block. Both are given by 
:
In this case, the rope must be able to support both the force of gravity and the force required to accelerate the block. Both are given by
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A block which weighs 
experiences a resultant force of 
that increases its velocity by 
. How long did this force act on the block?
A block which weighs 
Let's first write down the information we are given:
We know that 
. This equation can simplified to 
since acceleration is equal to velocity over time so 
.
Let's manipulate the equation by solving for time so we can simply plug in the numbers we are given:
so: 
=
which is our final answer.
Let's first write down the information we are given:
We know that
Let's manipulate the equation by solving for time so we can simply plug in the numbers we are given:
so:
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A 
object experiences a force of 
in its direction of motion for 
. What is the change in the velocity the object moved at during this time?
A
Let's first write down the information we are given:
We know that 
. This equation can simplified to 
since acceleration is equal to velocity over time so 
.
Let's manipulate the equation by solving for velocity so we can simply plug in the numbers we are given:
so: 
which is our final answer.
Let's first write down the information we are given:
We know that
Let's manipulate the equation by solving for velocity so we can simply plug in the numbers we are given:
so:
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A 
experiences a net force that increases its velocity by 
from rest for 
. What is the magnitude of this force?
A
Let's write down the information we are given:
We know that 
. This equation can simplified to 
since acceleration is equal to velocity over time so 
.
So we can just plug in the information we are given to solve for the Force:
, which is our final answer.
Let's write down the information we are given:
We know that
So we can just plug in the information we are given to solve for the Force:
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An object is pushed from rest with a force of 
that increases its velocity by 
for 
. What is the mass of this object?
An object is pushed from rest with a force of
Let's write down the information we are given:
We know that 
. This equation can simplified to 
since acceleration is equal to velocity over time so 
.
So we can solve for 
in this equation:
, which is our final answer.
Let's write down the information we are given:
We know that
So we can solve for
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A 
bowling ball is dropped from a height of 
onto a spring. If the spring is to compressed 
before halting the ball, what is the spring constant 
of the spring?
A
This questions requires an understanding of the conversion of gravitational potential energy to potential energy stored in the compression of a spring. The ball's initial potential energy is 
where 
is the mass of the ball, 
is the acceleration due to gravity, and 
is the height of the ball. The energy stored in a compressed spring is 
where 
is the potential energy, 
is the spring constant (typically given in 
) and 
is the compression of the spring. By setting the gravitational potential energy equal to the energy stored in the spring, we can solve for the spring constant 
:
This questions requires an understanding of the conversion of gravitational potential energy to potential energy stored in the compression of a spring. The ball's initial potential energy is
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A block with mass 
is attached to a vertically hanging spring with negligible mass and spring constant 
and is allowed to reach a new equilibrium position. Approximately how far does the spring stretch from its initial equilibrium position?
A block with mass
The new equilibrium position will be where the spring force is opposite in direction and equal in magnitude to the gravitational force on the mass. This occurs where 
where 
is the displacement of the spring from equilibrium. Solving for 
, we get 
.
The new equilibrium position will be where the spring force is opposite in direction and equal in magnitude to the gravitational force on the mass. This occurs where
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Supposed that an object attached to a spring with a spring constant of 
is displaced by a certain amount. The elastic potential energy contained within the spring is 
. What is the magnitude of the force acting on the object from the spring at this instant?
Supposed that an object attached to a spring with a spring constant of
The first step in answering this question is to recognize that we're dealing with two components of a spring; force and energy. Hence, we'll need to utilize both equations. Because both of them share the term for the displacement of the object, that will be the variable that links the two expressions.
Having isolated the variable for displacement, we can plug this value into the force equation.
Now, we can plug in the values given to us in the question stem.
Note that the negative sign shown here indicates the direction of the force. This is because the force is acting in the direction opposite the displacement of the object. Since the question stem asks for the magnitude of force, we don't need to include the negative sign in our answer.
The first step in answering this question is to recognize that we're dealing with two components of a spring; force and energy. Hence, we'll need to utilize both equations. Because both of them share the term for the displacement of the object, that will be the variable that links the two expressions.
Having isolated the variable for displacement, we can plug this value into the force equation.
Now, we can plug in the values given to us in the question stem.
Note that the negative sign shown here indicates the direction of the force. This is because the force is acting in the direction opposite the displacement of the object. Since the question stem asks for the magnitude of force, we don't need to include the negative sign in our answer.
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Suppose that an object of a certain mass that is hanging from a wall is suspended by two ropes. Which of the following gives an expression for the tension in each rope?
Suppose that an object of a certain mass that is hanging from a wall is suspended by two ropes. Which of the following gives an expression for the tension in each rope?
In this question, we're told that an object is hanging by being supported by two ropes. We're asked to find the tension in each rope.
When answering this question, we have to keep in mind all of the forces acting on the object. Since the object isn't moving, there is no net force acting on it. Moreover, we know that its weight acts downward while the tension in the ropes act upwards.
In this question, we're told that an object is hanging by being supported by two ropes. We're asked to find the tension in each rope.
When answering this question, we have to keep in mind all of the forces acting on the object. Since the object isn't moving, there is no net force acting on it. Moreover, we know that its weight acts downward while the tension in the ropes act upwards.
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An object weighing 
is hanging from a string with no mass and a length of 
. If this object is released from some initial height and reaches a maximum velocity of 
, what is the maximum amount of tension in the string at any point during the object's journey?
An object weighing
For this question, we're told that an object hanging from a string is released from an initial height and allowed to swing. We're given the maximum velocity, the object's mass, and the length of the string. We're asked to find the maximum tension in the string.
The first step towards solving this problem is to realize that we're dealing with centripetal motion. When the object is released from an initial height, it begins to fall. But as it falls, the string exerts a tension on the mass and causes the mass to travel in a circular motion. Thus, the tension of the string is supplying the centripetal force.
Furthermore, it's important to remember the point at which the object will have maximum velocity. When the object was initially released from some initial height, all of the object's energy was in the form of gravitational potential energy. As it falls, that potential energy is converted into kinetic energy. Once the object reaches the very bottom of its path of motion, all of the potential energy it once had has been converted into kinetic energy at that instant. Since all of the energy is in the form of kinetic energy, and the object's mass remains constant, we know that this must be the point where the maximum velocity occurs.
Thus, we need to analyze the situation when the mass has swung to the very bottom of the path. At this point, the tension is providing an upward force. Moreover, the only downward force acting on the object is its weight due to gravity. The net force resulting from these two will result in the overall centripetal force on the object. We can put this into an equation as follows.
Since we're trying to solve for the tension in the string, we can isolate the 
term in the above expression.
For this question, we're told that an object hanging from a string is released from an initial height and allowed to swing. We're given the maximum velocity, the object's mass, and the length of the string. We're asked to find the maximum tension in the string.
The first step towards solving this problem is to realize that we're dealing with centripetal motion. When the object is released from an initial height, it begins to fall. But as it falls, the string exerts a tension on the mass and causes the mass to travel in a circular motion. Thus, the tension of the string is supplying the centripetal force.
Furthermore, it's important to remember the point at which the object will have maximum velocity. When the object was initially released from some initial height, all of the object's energy was in the form of gravitational potential energy. As it falls, that potential energy is converted into kinetic energy. Once the object reaches the very bottom of its path of motion, all of the potential energy it once had has been converted into kinetic energy at that instant. Since all of the energy is in the form of kinetic energy, and the object's mass remains constant, we know that this must be the point where the maximum velocity occurs.
Thus, we need to analyze the situation when the mass has swung to the very bottom of the path. At this point, the tension is providing an upward force. Moreover, the only downward force acting on the object is its weight due to gravity. The net force resulting from these two will result in the overall centripetal force on the object. We can put this into an equation as follows.
Since we're trying to solve for the tension in the string, we can isolate the
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A mechanic is able to exert a maximum torque of 
using a small wrench. The mechanic adds an attachment to the wrench that doubles the handle length. What is the maximum amount of torque the mechanic can exert with the new wrench?
A mechanic is able to exert a maximum torque of
The maximum amount of torque is the magnitude of the torque vector 
, which is maximized when the angle between the applied force and the displacement vector is 90 degrees. Then the torque is given by 
, where 
is the length of the wrench and 
is the magnitude of the force applied. Since the mechanic doesn't get any stronger, but the handle length increases, the torque doubles.
The maximum amount of torque is the magnitude of the torque vector
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An object weighing 
sits on a board at a distance 
away from a fulcrum. At what distance to the right of the fulcrum would an 
object need to be in order to make the board flat?
An object weighing
For this question, we'll need to take torque into account. The amount of torque caused by the object to the left of the fulcrum needs to be equal to the torque of the object to the right of the fulcrum.
Moreover, we can write out the equation for torque.
Combining these equations, we can set an expression for the torque on the left and the torque on the right.
Then we can set them equal to each other.
Next, we isolate the term for the distance on the right and we cancel out common units.
Finally, we plug in the values given in the question stem. Because we're told that both objects are sitting flat on the board, we know the force of their weight is perpendicular to the board, which gives a value of 
.
For this question, we'll need to take torque into account. The amount of torque caused by the object to the left of the fulcrum needs to be equal to the torque of the object to the right of the fulcrum.
Moreover, we can write out the equation for torque.
Combining these equations, we can set an expression for the torque on the left and the torque on the right.
Then we can set them equal to each other.
Next, we isolate the term for the distance on the right and we cancel out common units.
Finally, we plug in the values given in the question stem. Because we're told that both objects are sitting flat on the board, we know the force of their weight is perpendicular to the board, which gives a value of
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Two small lead balls of masses 
and 
are fixed a distance 
apart on a laboratory bench. A student moves the larger mass toward the smaller mass so the distance is now only 
. After the mass was moved, by what factor did the gravitational force between the lead balls change?
Two small lead balls of masses
The gravitational force obeys an inverse-square law. Decreasing the distance by a factor of 
will increase the force by a factor of 
, or 
.
The gravitational force obeys an inverse-square law. Decreasing the distance by a factor of
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The gravitational force between two objects is 
. What would be the force of gravity between them if the mass of one of the objects is tripled and the distance between them is doubled?
The gravitational force between two objects is
So the fact that we don't actually know any of the numbers doesn't matter.
What matters is we have the equation 
The masses have a direct proportional relationship with the force. That means that if a mass doubles, the force doubles.
We can also see that the radius (distance between the masses) has an inverse square relationship with Force. So if we double the radius, we divide the force by 4.
So based on the problem, we triple a mass and double the radius. So we triple the force then divide by 4.
So the fact that we don't actually know any of the numbers doesn't matter.
What matters is we have the equation
The masses have a direct proportional relationship with the force. That means that if a mass doubles, the force doubles.
We can also see that the radius (distance between the masses) has an inverse square relationship with Force. So if we double the radius, we divide the force by 4.
So based on the problem, we triple a mass and double the radius. So we triple the force then divide by 4.
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Which of these equations is the universal gravitational equation?
Which of these equations is the universal gravitational equation?
is the equation for Universal gravitation.
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Which of these forces is identical to a normal force?
Which of these forces is identical to a normal force?
All of these are identical to a normal force.
All of these are identical to a normal force.
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