Newtonian Mechanics and Motion - MCAT Physical
Card 0 of 1869
A 50kg block resting on the ground experiences an upward acceleration of 4m/s2. What is the normal force acting on the block?
A 50kg block resting on the ground experiences an upward acceleration of 4m/s2. What is the normal force acting on the block?
The block experiences the force of gravity, plus the force of the upward acceleration


If the block is resting on the ground, then its total force must be zero, and the normal force must cancel out the net force above. The normal force is 300N.
The block experiences the force of gravity, plus the force of the upward acceleration
If the block is resting on the ground, then its total force must be zero, and the normal force must cancel out the net force above. The normal force is 300N.
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A 3kg book slides towards the right along a frictionless horizontal surface with initial velocity 5m/s, then suddenly encounters a long rough section with kinetic friction coefficient
. How far does the book travel along the rough surface before coming to rest? (Use
as needed)
A 3kg book slides towards the right along a frictionless horizontal surface with initial velocity 5m/s, then suddenly encounters a long rough section with kinetic friction coefficient . How far does the book travel along the rough surface before coming to rest? (Use
as needed)
We'll need to use the kinematic equation
to solve for d, the distance travelled when the book has stopped (
). Before solving for d, we need to calculate the acceleration caused by the frictional force, by using the following steps.
-
Find the normal force on the book,
.
-
Plug this normal force into
to solve for frictional force.
-
Find the acceleration caused by this frictional force, with
.
Step 1 gives
, so in step 2,
, giving an acceleration of
to the left (which we will define to be the negative horizontal direction).
Returning to the original kinematic equation,
.
Rearranging to solve for d gives 
We'll need to use the kinematic equation to solve for d, the distance travelled when the book has stopped (
). Before solving for d, we need to calculate the acceleration caused by the frictional force, by using the following steps.
-
Find the normal force on the book,
.
-
Plug this normal force into
to solve for frictional force.
-
Find the acceleration caused by this frictional force, with
.
Step 1 gives , so in step 2,
, giving an acceleration of
to the left (which we will define to be the negative horizontal direction).
Returning to the original kinematic equation, .
Rearranging to solve for d gives
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A 2kg box is at the top of a ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.
Imagine that the net force on the box is 16.5N when sliding down the ramp. What is the coefficient of kinetic friction for the box?
A 2kg box is at the top of a ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.
Imagine that the net force on the box is 16.5N when sliding down the ramp. What is the coefficient of kinetic friction for the box?
Since the box is moving when the net force on the box is determined, we can calculate the coefficient of kinetic friction for the box. The first step is determining what the net force on the box would be in the absence of friction. The net force on the box is given by the equation
.

The difference between the frictionless net force and the net force with friction is 0.8N. This means that the force of kinetic friction on the box is 0.8N, acting opposite the direction of motion. Knowing this, we can solve for the coefficient of kinetic friction using the equation 



Since the box is moving when the net force on the box is determined, we can calculate the coefficient of kinetic friction for the box. The first step is determining what the net force on the box would be in the absence of friction. The net force on the box is given by the equation .
The difference between the frictionless net force and the net force with friction is 0.8N. This means that the force of kinetic friction on the box is 0.8N, acting opposite the direction of motion. Knowing this, we can solve for the coefficient of kinetic friction using the equation
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What is the coefficient of kinetic friction of a 500g book sliding along a floor if the force of friction on the book is 4N?
What is the coefficient of kinetic friction of a 500g book sliding along a floor if the force of friction on the book is 4N?
For formula for the force of friction is
. We can rearrange this equation to solve for the coefficient of friction.

Remember that the normal force is equal to the force of gravity. Now we can plug in our given values and solve.

Don't forget to convert 500g to 0.5kg. The units cancel out, leaving the answer without any unit.
For formula for the force of friction is . We can rearrange this equation to solve for the coefficient of friction.
Remember that the normal force is equal to the force of gravity. Now we can plug in our given values and solve.
Don't forget to convert 500g to 0.5kg. The units cancel out, leaving the answer without any unit.
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A
object is originally at rest on an inclined plane, which forms an angle with the ground of
. If the coefficient of static friction is
, what is the force of friction that must be overcome for the object to begin moving?
A object is originally at rest on an inclined plane, which forms an angle with the ground of
. If the coefficient of static friction is
, what is the force of friction that must be overcome for the object to begin moving?
In this situation, we must simply remember how to calculate friction. Once the force of gravity overcomes the force of static friction, the object will slide. Our formula for friction is:

This means the force of friction is equal to the friction coefficient times the normal force. On an incline, the normal force is equal to the force of gravity times the cosine of the angle:


We can combine our formulas to give the force of friction.

Using the given coefficient of friction, mass, and angle, we can calculate the force of friction.


In this situation, we must simply remember how to calculate friction. Once the force of gravity overcomes the force of static friction, the object will slide. Our formula for friction is:
This means the force of friction is equal to the friction coefficient times the normal force. On an incline, the normal force is equal to the force of gravity times the cosine of the angle:
We can combine our formulas to give the force of friction.
Using the given coefficient of friction, mass, and angle, we can calculate the force of friction.
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When the force applied to a moving object is equal and opposite the force of kinetic friction, what happens to the object?
When the force applied to a moving object is equal and opposite the force of kinetic friction, what happens to the object?
It is important to understand the difference between static and kinetic friction. When an object is at rest, it takes more force to get it to start moving than to keep it moving. If you match the amount of static friction that can be generated when the object is at rest, it will not move because there is zero net force; the force applied must be greater than the static friction in order to initiate motion. Once the object begins moving, the force required to keep it moving decreases. If you match the force of kinetic friction, the object moves at a constant velocity because there is again no net force. Any more force will cause acceleration, while any less will cause deceleration.
It is important to understand the difference between static and kinetic friction. When an object is at rest, it takes more force to get it to start moving than to keep it moving. If you match the amount of static friction that can be generated when the object is at rest, it will not move because there is zero net force; the force applied must be greater than the static friction in order to initiate motion. Once the object begins moving, the force required to keep it moving decreases. If you match the force of kinetic friction, the object moves at a constant velocity because there is again no net force. Any more force will cause acceleration, while any less will cause deceleration.
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Which of the following could not influence the magnitude of frictional force acting on a book sliding across a horizontal table?
Which of the following could not influence the magnitude of frictional force acting on a book sliding across a horizontal table?
Frictional force is given by the equation:

The only factors that can change frictional force are the coefficient of friction, which is determined by the materials of the surfaces in contact, and the normal force. Since the normal force must have a magnitude such that the sum of forces perpendicular to the table equals zero, both the mass of the book and any external vertical forces would influence the normal force, and thus also would influence the frictional force.

Frictional force is given by the equation:
The only factors that can change frictional force are the coefficient of friction, which is determined by the materials of the surfaces in contact, and the normal force. Since the normal force must have a magnitude such that the sum of forces perpendicular to the table equals zero, both the mass of the book and any external vertical forces would influence the normal force, and thus also would influence the frictional force.
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A
block rests on a wooden table. What is the force of the table upon the block?
A block rests on a wooden table. What is the force of the table upon the block?
In this question, the net force on the block is zero. We know from Newton's second law that any non-zero force will produce an acceleration, resulting in movement of some sort. Since the block is at rest, and not moving, we can conclude that the net force is zero.

The forces acting on the block are the force of gravity, pulling the block downward, and the normal force, pushing the block upward. The force of the block on the table will be the force from gravity, while the force of the table on the block will be the normal force. Since these are the only two forces acting on the block, we can add them together to get the net force.

Reorganizing the equation, we can set the two forces equal. This is a reflection of Newton's third law.

Gravitational force is equal to the mass of the object times the acceleration from gravity.

Using these values, given in the question, we can find the normal force, or the force of the table on the block.



The final normal force is positive because it acts in the upward direction, opposite of gravity.
In this question, the net force on the block is zero. We know from Newton's second law that any non-zero force will produce an acceleration, resulting in movement of some sort. Since the block is at rest, and not moving, we can conclude that the net force is zero.
The forces acting on the block are the force of gravity, pulling the block downward, and the normal force, pushing the block upward. The force of the block on the table will be the force from gravity, while the force of the table on the block will be the normal force. Since these are the only two forces acting on the block, we can add them together to get the net force.
Reorganizing the equation, we can set the two forces equal. This is a reflection of Newton's third law.
Gravitational force is equal to the mass of the object times the acceleration from gravity.
Using these values, given in the question, we can find the normal force, or the force of the table on the block.
The final normal force is positive because it acts in the upward direction, opposite of gravity.
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A
block rests on a wooden table. A spring scale attached to the right side of the block is very gently pulled to the right with increasing force. Which of these is true, assuming a frictional force between the table and the block?
A block rests on a wooden table. A spring scale attached to the right side of the block is very gently pulled to the right with increasing force. Which of these is true, assuming a frictional force between the table and the block?
This is a classic experiment to determine the coefficient of static friction,
. The frictional interaction of the block on the table causes an equal and opposite force to be generated against the pulling on the spring scale. The coefficient of static friction represents the ratio of the horizontal forces caused by surface irregularities to the vertical force due to gravity.

When the frictional force is just barely overcome, the block will start moving to the right, but as soon as it does, the tension in the spring diminishes and static frictional forces prevail. The block thus starts moving and almost immediately stops.
This is a classic experiment to determine the coefficient of static friction, . The frictional interaction of the block on the table causes an equal and opposite force to be generated against the pulling on the spring scale. The coefficient of static friction represents the ratio of the horizontal forces caused by surface irregularities to the vertical force due to gravity.
When the frictional force is just barely overcome, the block will start moving to the right, but as soon as it does, the tension in the spring diminishes and static frictional forces prevail. The block thus starts moving and almost immediately stops.
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A
block rests on a wooden table. A spring scale attached to the right side of the block is very gently pulled to the right with increasing force. The block just begins to move when the spring scale reads
. What is the coefficient of static friction?
A block rests on a wooden table. A spring scale attached to the right side of the block is very gently pulled to the right with increasing force. The block just begins to move when the spring scale reads
. What is the coefficient of static friction?
The coefficient of friction is a ratio of the force initiating horizontal displacement of an object and the downwards force of the object due to gravity


Note that the units cancel since we are finding the ratio of two forces. Although it is possible to have values for the coefficient of static friction greater than one, these systems are highly unusual.
The coefficient of friction is a ratio of the force initiating horizontal displacement of an object and the downwards force of the object due to gravity
Note that the units cancel since we are finding the ratio of two forces. Although it is possible to have values for the coefficient of static friction greater than one, these systems are highly unusual.
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A
block rests on a wooden table with a coefficient of kinetic friction equal to
. A spring scale attached to the right side of the block is very gently pulled to the right with increasing force. If the spring scale is pulled with a force of
, what is the acceleration of the system?
A block rests on a wooden table with a coefficient of kinetic friction equal to
. A spring scale attached to the right side of the block is very gently pulled to the right with increasing force. If the spring scale is pulled with a force of
, what is the acceleration of the system?
The force of kinetic friction is given by the equation:

We can calculate the frictional force using the values from the question.


There are four force acting on the block: force from gravity, normal force, force of the spring scale, and force of friction. The normal force will be equal and opposite to the force of gravity, allowing us to cancel the forces in the vertical direction. This leaves us with a net force calculation for the horizontal force: the spring scale force and the force of friction. Note that the frictional force remains negative, as it acts in the opposite direction to the spring scale force.


Now that we know the net force and the mass of the block, we can calculate the acceleration using Newton's second law.



The force of kinetic friction is given by the equation:
We can calculate the frictional force using the values from the question.
There are four force acting on the block: force from gravity, normal force, force of the spring scale, and force of friction. The normal force will be equal and opposite to the force of gravity, allowing us to cancel the forces in the vertical direction. This leaves us with a net force calculation for the horizontal force: the spring scale force and the force of friction. Note that the frictional force remains negative, as it acts in the opposite direction to the spring scale force.
Now that we know the net force and the mass of the block, we can calculate the acceleration using Newton's second law.
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A certain planet has three times the radius of Earth and nine times the mass. How does the acceleration of gravity at the surface of this planet (ag) compare to the acceleration at the surface of Earth (g)?
A certain planet has three times the radius of Earth and nine times the mass. How does the acceleration of gravity at the surface of this planet (ag) compare to the acceleration at the surface of Earth (g)?
The acceleration of gravity is given by the equation
, where G is constant.
For Earth,
.
For the new planet,
.
So, the acceleration is the same in both cases.
The acceleration of gravity is given by the equation , where G is constant.
For Earth, .
For the new planet,
.
So, the acceleration is the same in both cases.
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A hypothetical planet has a radius equal to twice that of Earth, with the same mass as Earth. How much would a person weigh on the surface of this hypothetical planet if they weighed 1000N on earth?
A hypothetical planet has a radius equal to twice that of Earth, with the same mass as Earth. How much would a person weigh on the surface of this hypothetical planet if they weighed 1000N on earth?
The force due to gravity on any object can be given by the equation below.

is the gravitational constant,
is the mass of the earth,
is the mass of the object, and
is the distance between the center of each object.
In our question, the only value to change is the radius of the new planet; both masses and
remain constant. The effect of doubling the radius on the force is given below.

The person's weight on the new planet would be one-fourth their weight on Earth.

The force due to gravity on any object can be given by the equation below.
is the gravitational constant,
is the mass of the earth,
is the mass of the object, and
is the distance between the center of each object.
In our question, the only value to change is the radius of the new planet; both masses and remain constant. The effect of doubling the radius on the force is given below.
The person's weight on the new planet would be one-fourth their weight on Earth.
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A person stands on a scale in an elevator. When the elevator is moving upwards at a constant velocity of 0.5m/s, the scale reads 500N. If the elevator then slows down, with a deceleration of 0.5m/s2, what is the new reading?
A person stands on a scale in an elevator. When the elevator is moving upwards at a constant velocity of 0.5m/s, the scale reads 500N. If the elevator then slows down, with a deceleration of 0.5m/s2, what is the new reading?
It may be helpful to start with a free-body diagram showing the forces acting on the person. We have the gravitational force, Fg, downwards, and the normal force of the scale, Fn, upwards.
Use these to write the net force equation.

First we need to solve for the person’s mass, m. When the elevator is moving at a constant rate, we are given that Fn = 500N, and we can solve for m.
.

When the elevator comes to rest, then we have an acceleration of -0.5 m/s2 (downwards acceleration). Plugging this in, we can solve for the new Fn.


It may be helpful to start with a free-body diagram showing the forces acting on the person. We have the gravitational force, Fg, downwards, and the normal force of the scale, Fn, upwards.
Use these to write the net force equation.
First we need to solve for the person’s mass, m. When the elevator is moving at a constant rate, we are given that Fn = 500N, and we can solve for m.
.
When the elevator comes to rest, then we have an acceleration of -0.5 m/s2 (downwards acceleration). Plugging this in, we can solve for the new Fn.
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What is the tension in a rope that is used to pull a
box straight upwards with an acceleration of
?
What is the tension in a rope that is used to pull a box straight upwards with an acceleration of
?
For the problem we need to understand Newton's second law:
. The net upward forces and net downward forces must equal the product of mass and acceleration. Since the box is traveling upwards with an acceleration of 1m/s2 we can indicate upward forces as positive and downward forces as negative. Indicating T as tension and Fg as the weight of the box, we can find the value of tension with the equation
.

For the problem we need to understand Newton's second law: . The net upward forces and net downward forces must equal the product of mass and acceleration. Since the box is traveling upwards with an acceleration of 1m/s2 we can indicate upward forces as positive and downward forces as negative. Indicating T as tension and Fg as the weight of the box, we can find the value of tension with the equation
.
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A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

What is the apparent weight of the weights on the platform as they fall after the rope has been cut?
A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
What is the apparent weight of the weights on the platform as they fall after the rope has been cut?
First, we need to determine what the apparent weight actually is. If we draw a free body diagram of the weights, we note that the force due to gravity acts downward and the force of the platform on the weights (the normal force) acts upward.

The normal force, FN in the above diagram, is what a scale reads, and is thus the apparent weight. Finding the normal force is not as easy as equating it to the weight because the system is in free-fall, and thus accelerating. We can use Newton’s second law and equate it to acceleration to find the normal force.
Assuming downward is the positive y direction, we can solve for FN.
mg – FN = ma
FN = mg – ma = 0 N
Logically, this makes sense. If an object is in free fall, the acceleration of the system is the same as the acceleration due to gravity.
First, we need to determine what the apparent weight actually is. If we draw a free body diagram of the weights, we note that the force due to gravity acts downward and the force of the platform on the weights (the normal force) acts upward.
The normal force, FN in the above diagram, is what a scale reads, and is thus the apparent weight. Finding the normal force is not as easy as equating it to the weight because the system is in free-fall, and thus accelerating. We can use Newton’s second law and equate it to acceleration to find the normal force.
Assuming downward is the positive y direction, we can solve for FN.
mg – FN = ma
FN = mg – ma = 0 N
Logically, this makes sense. If an object is in free fall, the acceleration of the system is the same as the acceleration due to gravity.
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Planet X has mass
and radius
, and the gravitational acceleration on its surface is
. What is the gravitational acceleration at the location of a satellite orbiting at a distance
from the surface of planet X?
Planet X has mass and radius
, and the gravitational acceleration on its surface is
. What is the gravitational acceleration at the location of a satellite orbiting at a distance
from the surface of planet X?
Gravitational acceleration is related to distance via the equation:

is the gravitational constant,
is the mass of the attracting object, and
is the distance from its center.
In this case, the initial distance (on the surface) is
from the center of planet X, and the distance of the satellite is
from the center. We know that gravitational acceleration is proportional to the distance squared, and we know the acceleration at the surface. Using these values, we can solve for the acceleration on the satellite.



By expanding the equation for the acceleration on the satellite, we can see that it is equal to one-sixteenth the acceleration at the surface, based on our original equation. Substitute the value of the surface acceleration to get the final answer.


Gravitational acceleration is related to distance via the equation:
is the gravitational constant,
is the mass of the attracting object, and
is the distance from its center.
In this case, the initial distance (on the surface) is from the center of planet X, and the distance of the satellite is
from the center. We know that gravitational acceleration is proportional to the distance squared, and we know the acceleration at the surface. Using these values, we can solve for the acceleration on the satellite.
By expanding the equation for the acceleration on the satellite, we can see that it is equal to one-sixteenth the acceleration at the surface, based on our original equation. Substitute the value of the surface acceleration to get the final answer.
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An astronaut standing on a scale on the moon observes that he weighs
. If the acceleration due to gravity on the moon's is one-sixth of its value on Earth's surface, what is the astronaut's mass on the earth?
An astronaut standing on a scale on the moon observes that he weighs . If the acceleration due to gravity on the moon's is one-sixth of its value on Earth's surface, what is the astronaut's mass on the earth?
To relate force and mass, we use Newton's second law:

We are given his weight (force) on the moon, and we are told the relative gravitational acceleration on the moon. Using these values, we can find the astronaut's mass.





Since mass is the same regardless of gravitational acceleration, this is the same as the astronaut's mass on the earth.
To relate force and mass, we use Newton's second law:
We are given his weight (force) on the moon, and we are told the relative gravitational acceleration on the moon. Using these values, we can find the astronaut's mass.
Since mass is the same regardless of gravitational acceleration, this is the same as the astronaut's mass on the earth.
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A
block lies upon a frictionless surface. A string is attached to the right side of the block, passed over a pulley, and then attached to a
mass suspended by the string. When the block is held still, what is the tension in the string?
A block lies upon a frictionless surface. A string is attached to the right side of the block, passed over a pulley, and then attached to a
mass suspended by the string. When the block is held still, what is the tension in the string?
The block cannot provide any frictional force to the left because it is resting on a frictionless table. Whether the block is allowed to move or not, the tension force in the string is all generated by the suspended
mass. The force in the string is equal and opposite to the gravitational force acting on the suspended mass.


The system is in equilibrium when no parts are in motion. Alternatively stated, the upward tension in the string is equal and opposite to the downwards force on the mass. If the block is not allowed to move as more mass is added, at some point the tensile force in the string will exceed its mechanical limits, and the string will break.
The block cannot provide any frictional force to the left because it is resting on a frictionless table. Whether the block is allowed to move or not, the tension force in the string is all generated by the suspended mass. The force in the string is equal and opposite to the gravitational force acting on the suspended mass.
The system is in equilibrium when no parts are in motion. Alternatively stated, the upward tension in the string is equal and opposite to the downwards force on the mass. If the block is not allowed to move as more mass is added, at some point the tensile force in the string will exceed its mechanical limits, and the string will break.
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A ball is thrown vertically with an initial velocity,
, and returns to its original position after time
. How would the value of
be affected if the ball were thrown in the same manner on the moon, where gravitational acceleration is one-sixth the gravitational acceleration on Earth?
A ball is thrown vertically with an initial velocity, , and returns to its original position after time
. How would the value of
be affected if the ball were thrown in the same manner on the moon, where gravitational acceleration is one-sixth the gravitational acceleration on Earth?
To solve this question, we will need to use the equation for acceleration:

In this case, the initial velocity will be equal to the final velocity, but opposite in magnitude. The initial velocity is in the upward direction, while the final velocity is downward.


Plug this value into the equation for acceleration.

The velocity value is constant, regardless of the planet. Substitute the acceleration for each planet to determine the change in the time variable.


We can see that the time of flight on the moon is equal to six times the time of flight on Earth.
To solve this question, we will need to use the equation for acceleration:
In this case, the initial velocity will be equal to the final velocity, but opposite in magnitude. The initial velocity is in the upward direction, while the final velocity is downward.
Plug this value into the equation for acceleration.
The velocity value is constant, regardless of the planet. Substitute the acceleration for each planet to determine the change in the time variable.
We can see that the time of flight on the moon is equal to six times the time of flight on Earth.
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