Specific Forces - High School Physics
Card 0 of 20
A block is on a smooth horizontal surface (
), connected by a tine cord that passes over a pulley to a second block which hangs vertically (
). Find the formula for the acceleration of the system. Ignore friction and the masses of the pulley and cord.
A block is on a smooth horizontal surface (
To determine the net force on a system, it is important to consider all the forces between each object in the same direction. The direction to be considered is along the line of the rope. The system is considered to be all the blocks together as they will all move with the same acceleration.
According to Newton’s 3rd Law, the Tension in the rope caused by A pulling on B and B pulling on A is equal in magnitude and opposite in direction. Therefore these two Tension forces cancel out.
From Newton’s 2nd law we know that the net force on a system is equal to the mass of the system times the acceleration of the system
Therefore
We can calculate the force of gravity from the equation Fg = mg
Therefore
We can rearrange this equation and solve for a
To determine the net force on a system, it is important to consider all the forces between each object in the same direction. The direction to be considered is along the line of the rope. The system is considered to be all the blocks together as they will all move with the same acceleration.
According to Newton’s 3rd Law, the Tension in the rope caused by A pulling on B and B pulling on A is equal in magnitude and opposite in direction. Therefore these two Tension forces cancel out.
From Newton’s 2nd law we know that the net force on a system is equal to the mass of the system times the acceleration of the system
Therefore
We can calculate the force of gravity from the equation Fg = mg
Therefore
We can rearrange this equation and solve for a
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Three blocks on a frictionless horizontal surface as in contact with each other. A force is applied to the first block (
). Determine the net force on block B terms of 
, 
, and 
and 
.
Three blocks on a frictionless horizontal surface as in contact with each other. A force is applied to the first block (
To determine the net force on a system, it is important to consider all the forces between each object in the horizontal direction. The system is considered to be all the blocks together as they will all move with the same acceleration.
According to Newton’s 3rd Law 
and 
are equal in magnitude and opposite in direction so they will cancel out. The same goes for 
and 
.
Therefore our equation is reduce to
From Newton’s 2nd law we know that the net force on a system is equal to the mass of the system times the acceleration of the system
Therefore
This means that the acceleration on any one block is
So when solving for the net force on any one block, the net force on the single block will be equal to that block's mass times the acceleration of the system. Therefore for block B, the net force is
To determine the net force on a system, it is important to consider all the forces between each object in the horizontal direction. The system is considered to be all the blocks together as they will all move with the same acceleration.
According to Newton’s 3rd Law 
Therefore our equation is reduce to
From Newton’s 2nd law we know that the net force on a system is equal to the mass of the system times the acceleration of the system
Therefore
This means that the acceleration on any one block is
So when solving for the net force on any one block, the net force on the single block will be equal to that block's mass times the acceleration of the system. Therefore for block B, the net force is
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A child slides down a slide with a 30 degree incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.
A child slides down a slide with a 30 degree incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.
Consider the net forces acting on the object causing it to accelerate.
To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction. Using trigonometric functions we get that
We know that the force of gravity is equal to mg
According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.
The force of friction is directly related to 
(the coefficient of friction) times the normal force. In this case the normal force is equal to the y component of the force of gravity.
Therefore
If we substitute this in our original net force equation
Notice that mass is in each piece of the equation so we can cancel it out.
We can also use our kinematic equations to determine the speed of the object at the bottom of the incline. We can represent 
to be the length of the slope.
Since the ball is assumed to be at rest at the top of the incline, the initial velocity at the top will be 0.
Therefore
Solving for 
on its own we get
In the original situation, the force of gravity is the only force pulling on the object. Therefore the acceleration is
We can substitute this value into our velocity equation
In the second situation the velocity is one half of the original velocity. Therefore
We can place this back into our kinematic equation and rearrange it to solve for the acceleration.
We can then substitute that into our net force acceleration equation
Notice that g is in all of these terms so we can cancel it out.
Now we can substitute and solve
Consider the net forces acting on the object causing it to accelerate.
To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction. Using trigonometric functions we get that
We know that the force of gravity is equal to mg
According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.
The force of friction is directly related to 
Therefore
If we substitute this in our original net force equation
Notice that mass is in each piece of the equation so we can cancel it out.
We can also use our kinematic equations to determine the speed of the object at the bottom of the incline. We can represent 
Since the ball is assumed to be at rest at the top of the incline, the initial velocity at the top will be 0.
Therefore
Solving for 
In the original situation, the force of gravity is the only force pulling on the object. Therefore the acceleration is
We can substitute this value into our velocity equation
In the second situation the velocity is one half of the original velocity. Therefore
We can place this back into our kinematic equation and rearrange it to solve for the acceleration.
We can then substitute that into our net force acceleration equation
Notice that g is in all of these terms so we can cancel it out.
Now we can substitute and solve
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Susan is trying to push a 
crate across the floor. She observes that the force of friction between the crate and the floor is 
. What is the coefficient of static friction? Assume 
Susan is trying to push a 
The equation for the force of friction is 
, where μ is the coefficient of static friction.
The normal force is equal to the mass times acceleration due to gravity, but in the opposite direction (negative of the force of gravity).
Since the problem tells us that the force due to friction is 
, we can plug these values into our original equation to solve for the coefficient of friction.
The coefficient of friction has no units.
The equation for the force of friction is 
The normal force is equal to the mass times acceleration due to gravity, but in the opposite direction (negative of the force of gravity).
Since the problem tells us that the force due to friction is 
The coefficient of friction has no units.
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A 
crate slides across a floor for 
before coming to rest 
from its original position. What is the coefficient of kinetic friction on the crate? Assume 
A 
The equation for the force due to friction is 
, where μ is the coefficient of kinetic friction. Since there is only one force acting upon the object, the force due to friction, we can find its value using the equation 
. We can equate these two force equations, meaning that 
. We can solve for the normal force, but we need to find ma in order to find 
.
The problem gives us the mass of the crate, but we have to solve for the acceleration.
Start by finding the initial velocity. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.
Plug in our given values and solve.
We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.
Plug in the given values to solve for acceleration.
Now that we have the acceleration and the mass, we can return to our first equation for force.
The normal force is the same as the mass times gravity.
In this format, the masses cancel on both sides of the equation/
Now we can plug in our value for acceleration and gravity to solve for the coefficient of friction.
The equation for the force due to friction is 
The problem gives us the mass of the crate, but we have to solve for the acceleration.
Start by finding the initial velocity. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.
Plug in our given values and solve.
We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.
Plug in the given values to solve for acceleration.
Now that we have the acceleration and the mass, we can return to our first equation for force.
The normal force is the same as the mass times gravity.
In this format, the masses cancel on both sides of the equation/
Now we can plug in our value for acceleration and gravity to solve for the coefficient of friction.
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A 
box is released on a 25 degree incline and accelerates down the ramp at 
. What is the coefficient of kinetic friction impeding its motion?
A 
Consider the net forces acting on the object causing it to accelerate.
To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction. Using trigonometric functions we get that
We know that the force of gravity is equal to mg
According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.
The force of friction is directly related to μ (the coefficient of friction) times the normal force. In this case the normal force is equal to the y component of the force of gravity.
Therefore
If we substitute this in our original net force equation
Notice that mass is in each piece of the equation so we can cancel it out.
Now we can rearrange and solve for the coefficient of friction
Consider the net forces acting on the object causing it to accelerate.
To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction. Using trigonometric functions we get that
We know that the force of gravity is equal to mg
According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.
The force of friction is directly related to μ (the coefficient of friction) times the normal force. In this case the normal force is equal to the y component of the force of gravity.
Therefore
If we substitute this in our original net force equation
Notice that mass is in each piece of the equation so we can cancel it out.
Now we can rearrange and solve for the coefficient of friction
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Which of the following cannot be true of an object on a given surface?
Which of the following cannot be true of an object on a given surface?
Kinetic friction is never greater than static friction. More force is always requires to overcome static friction than is required to overcome kinetic friction. It can require a large force to initiate motion, causing an initial acceleration by overcoming static friction. Once motion has begun, however, less for is required to maintain the motion due to the principles of Newton's first law and inertia.
Kinetic friction is never greater than static friction. More force is always requires to overcome static friction than is required to overcome kinetic friction. It can require a large force to initiate motion, causing an initial acceleration by overcoming static friction. Once motion has begun, however, less for is required to maintain the motion due to the principles of Newton's first law and inertia.
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Two toy cars (15kg and 3kg) are released simultaneously on an inclined plane that makes an angle of 28 degrees with the horizontal. Which statement best describes their acceleration once they are released. Assume that the track is frictionless.
Two toy cars (15kg and 3kg) are released simultaneously on an inclined plane that makes an angle of 28 degrees with the horizontal. Which statement best describes their acceleration once they are released. Assume that the track is frictionless.
Since the only force pulling both cars down the slope is the force of gravity, both cars will accelerate at the same rate.
It is possible to prove this mathematically by examining the x-component of the gravitational pull.
We know that the force of gravity is equal to mg
According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.
Both sides have mass so this cancels out of the equation.
Since this is independent of mass, this will be true for both objects.
Since the only force pulling both cars down the slope is the force of gravity, both cars will accelerate at the same rate.
It is possible to prove this mathematically by examining the x-component of the gravitational pull.
We know that the force of gravity is equal to mg
According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.
Both sides have mass so this cancels out of the equation.
Since this is independent of mass, this will be true for both objects.
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A horizontal force accelerates a box from rest across a horizontal surface where friction is present at a constant rate. Then the experiment is repeated. All conditions remain the same with the exception that the horizontal force is doubled. What happens to the box’s acceleration.
A horizontal force accelerates a box from rest across a horizontal surface where friction is present at a constant rate. Then the experiment is repeated. All conditions remain the same with the exception that the horizontal force is doubled. What happens to the box’s acceleration.
According to Newton’s 2nd Law the net force on the object is equal to the mass times the acceleration. The net force is also equal to the sum of the forces involved.
In the second experiment the horizontal applied force is doubled. However, this does not have any effect on the friction on the object. Therefore the new equation is
From this equation we can see that the new acceleration will be more than double the original value.
To examine this numerically, let’s assume that the mass of the box is 
, the friction force is 
and the applied force originally was 
.
Then, let’s double the horizontal force, keeping the friction force the same.
This value is more than double the original value, confirming our answer.
According to Newton’s 2nd Law the net force on the object is equal to the mass times the acceleration. The net force is also equal to the sum of the forces involved.
In the second experiment the horizontal applied force is doubled. However, this does not have any effect on the friction on the object. Therefore the new equation is
From this equation we can see that the new acceleration will be more than double the original value.
To examine this numerically, let’s assume that the mass of the box is 
Then, let’s double the horizontal force, keeping the friction force the same.
This value is more than double the original value, confirming our answer.
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Jerry wants to lift a 
ball with exactly enough force so that its upward velocity is constant. How much force should he use? Assume 
Jerry wants to lift a 
If the velocity on an object is constant, that means it has no acceleration. If it has no acceleration, that means that the net force on the object is equal to zero. We can see this conclusion by using Newton's second law.
Another way to think of 
is the sum of all the forces. Since the only two forces acting upon the ball are gravity and Jerry's lifting force, we can see: 
.
Since the net force is zero, the magnitude of Jerry's force must equal the magnitude of the force of gravity, but in the opposite direction.
This means that once we find 
, then Jerry's lifting force will be the same magnitude but in the opposite direction. Use Newton's second law to find the force of gravity.
This means that Jerry's lifting force will be 
.
If the velocity on an object is constant, that means it has no acceleration. If it has no acceleration, that means that the net force on the object is equal to zero. We can see this conclusion by using Newton's second law.
Another way to think of 
Since the net force is zero, the magnitude of Jerry's force must equal the magnitude of the force of gravity, but in the opposite direction.
This means that once we find 
This means that Jerry's lifting force will be 
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What is the acceleration of two falling skydivers (total mass = 
including parachute) when the upward force of air resistance is equal to one fourth of their weight?
What is the acceleration of two falling skydivers (total mass = 
There are two forces acting on the falling skydivers. The first is the force of gravity (their weight) which we will denote as 
. The second is the force of air resistance pushing up against them which we will denote as 
.
Newton’s 2nd Law tells us that the net force acting on these two objects is equal to their mass times their acceleration.
The net force is the sum of both forces acting together.
We know that the air resistance acts opposite to gravity and is equal to ¼ their total weight.
Substitute this value into our net force equation.
This simplifies to
The force of gravity is calculated by multiplying the mass by the acceleration due to gravity.
Substitute in our known variables and solve
There are two forces acting on the falling skydivers. The first is the force of gravity (their weight) which we will denote as 
Newton’s 2nd Law tells us that the net force acting on these two objects is equal to their mass times their acceleration.
The net force is the sum of both forces acting together.
We know that the air resistance acts opposite to gravity and is equal to ¼ their total weight.
Substitute this value into our net force equation.
This simplifies to
The force of gravity is calculated by multiplying the mass by the acceleration due to gravity.
Substitute in our known variables and solve
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Sally is to walk across a “high wire” that has been strung horizontally between two buildings that are 
apart. The sag (dip) in the rope when she stands at the midpoint is 
. If her mass is 
, what is the tension in the rope at this point?
Sally is to walk across a “high wire” that has been strung horizontally between two buildings that are 
The first thing is to identify the forces involved in this situation. There is the force of gravity (or her weight) which is pulling down on the rope. We can calculate this by
The other forces are the force of Tension on each side of the wire as she stands in the midpoint. These two Tension forces are what hold up Sally and keep her from falling. However, these two Tensions forces are at an 
angle below the horizontal. This means that we need to analyze the components of the Tension force. The 
-components of each Tension force are equal in magnitude and opposite in direction as this is what keeps the rope connected to both buildings. The 
-components of each Tension force are equal in magnitude and in the same direction as they both are keeping Sally up. So we can sum up the forces acting in the 
-direction as:
Which can be simplified to
Since Sally is not accelerating, the forces are balanced and the net force must equal 
.
Earlier we calculated the force of gravity so we can substitute this in to find the Tension in the 
Direction.
This is the Tension in the 
-direction. However, the problem is asking for the overall Tension in the wire. At this point, we must use trigonometric functions to determine the hypotenuse (the overall Tension) in the wire. Since the 
-component of the Tension is the opposite side of the triangle from the 
angle, we can use cosine to find our hypotenuse.
The first thing is to identify the forces involved in this situation. There is the force of gravity (or her weight) which is pulling down on the rope. We can calculate this by
The other forces are the force of Tension on each side of the wire as she stands in the midpoint. These two Tension forces are what hold up Sally and keep her from falling. However, these two Tensions forces are at an 
Which can be simplified to
Since Sally is not accelerating, the forces are balanced and the net force must equal 
Earlier we calculated the force of gravity so we can substitute this in to find the Tension in the 
This is the Tension in the 
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Michael lands on a new planet. If his mass is 
and the acceleration due to gravity on this planet is 
, what would his weight be?
Michael lands on a new planet. If his mass is 
Weight is a particular force that is equal to mass times acceleration due to gravity. Start with Newton's second law, 
.
Plug in the given values to solve.
Weight is a particular force that is equal to mass times acceleration due to gravity. Start with Newton's second law, 
Plug in the given values to solve.
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If the mass of the object is 
and 
, what is the value of 
? Assume 
If the mass of the object is 
will be the weight of the object. Weight is a very specific force: it is the mass times gravity. As it turns out, the angle is irrelevant in finding weight.
Using Newton's second law and the given values for mass and gravity, we can solve for 
.
Note that the weight is negative, because it is acting in the downward direction.
Using Newton's second law and the given values for mass and gravity, we can solve for 
Note that the weight is negative, because it is acting in the downward direction.
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A 
ball falls off a cliff. What is the force of gravity on the ball? Assume 
A 
Newton's second law states:
In this case the acceleration will be the constant acceleration due to gravity on Earth.
Use the acceleration of gravity and the mass of the ball to solve for the force on the ball.
The answer is negative because the force is directed downward. Since gravity is always acting downward, a force due to gravity will always be negative.
Newton's second law states:
In this case the acceleration will be the constant acceleration due to gravity on Earth.
Use the acceleration of gravity and the mass of the ball to solve for the force on the ball.
The answer is negative because the force is directed downward. Since gravity is always acting downward, a force due to gravity will always be negative.
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The acceleration of gravity on the moon is significantly less than the acceleration of gravity on earth. What will happen to an astronaut's weight and mass on the moon, compared to her weight and mass on Earth?
The acceleration of gravity on the moon is significantly less than the acceleration of gravity on earth. What will happen to an astronaut's weight and mass on the moon, compared to her weight and mass on Earth?
Mass is a measure of how much matter is in an object, while weight is a measurement of the effective force of gravity on the object.
The amount of matter in the astronaut does not change; therefore, she has the same mass on the moon as she has on Earth.
Her weight, however, will change due to the change in gravity. The force of gravity will be the product of the acceleration and the astronaut's mass.
If the gravity is less on the moon than on Earth, then the force of gravity on the astronaut will also be less on the moon; thus, she will weigh less.
Mass is a measure of how much matter is in an object, while weight is a measurement of the effective force of gravity on the object.
The amount of matter in the astronaut does not change; therefore, she has the same mass on the moon as she has on Earth.
Her weight, however, will change due to the change in gravity. The force of gravity will be the product of the acceleration and the astronaut's mass.
If the gravity is less on the moon than on Earth, then the force of gravity on the astronaut will also be less on the moon; thus, she will weigh less.
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A woman stands on the edge of a cliff and drops two rocks, one of mass 
and one of 
, from the same height. Which one experiences the greater acceleration?
A woman stands on the edge of a cliff and drops two rocks, one of mass 
Even though the rocks have different masses, the acceleration on both will be 
, the acceleration due to gravity. We can look at Newton's second law to see the force experienced by the rocks:
When objects are in free-fall, the acceleration will be equal to the acceleration from gravity, regardless of the mass of the object.
Even though the rocks have different masses, the acceleration on both will be 
When objects are in free-fall, the acceleration will be equal to the acceleration from gravity, regardless of the mass of the object.
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A person stands on a scale in an elevator. His apparent weight will be the greatest when the elevator
A person stands on a scale in an elevator. His apparent weight will be the greatest when the elevator
Consider a person standing on a scale in an elevator that is not moving. The person is exerting a downward force onto the scale equal to their weight force. The scale exerts a force upward that is equal to the downward force of gravity. This upward force is the reading on the scale.
Now consider a person standing on a scale in an elevator that is moving at a constant velocity. The person is exerting a downward force onto the scale equal to the force of gravity. The scale exerts a force upward. Because the elevator is moving at a constant velocity, these forces are balanced, and the scale would read the same as if you were not moving.
Now consider a person standing in an elevator that is accelerating. There are still two forces, the downward force of gravity, and the upward normal force. Since the elevator is accelerating, these forces are unbalanced in the direction of the motion. If the elevator is accelerating down, gravity is larger than the normal force. If the elevator is accelerating up, the normal force is greater than gravity.
Since the scale reads the magnitude of the normal force, the time when the normal force is the greatest is when the elevator is accelerating in an upward direction.
Consider a person standing on a scale in an elevator that is not moving. The person is exerting a downward force onto the scale equal to their weight force. The scale exerts a force upward that is equal to the downward force of gravity. This upward force is the reading on the scale.
Now consider a person standing on a scale in an elevator that is moving at a constant velocity. The person is exerting a downward force onto the scale equal to the force of gravity. The scale exerts a force upward. Because the elevator is moving at a constant velocity, these forces are balanced, and the scale would read the same as if you were not moving.
Now consider a person standing in an elevator that is accelerating. There are still two forces, the downward force of gravity, and the upward normal force. Since the elevator is accelerating, these forces are unbalanced in the direction of the motion. If the elevator is accelerating down, gravity is larger than the normal force. If the elevator is accelerating up, the normal force is greater than gravity.
Since the scale reads the magnitude of the normal force, the time when the normal force is the greatest is when the elevator is accelerating in an upward direction.
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A 
weight and a 
weight are dropped simultaneously from the same height. Ignore air resistance. Compare their accelerations.
A 
No explanation available
No explanation available
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Susan is trying to push a 
crate across the floor. She observes that the force of friction between the crate and the floor is 
. What is the coefficient of static friction?
Susan is trying to push a
The equation for the force of friction is 
, where 
is the coefficient of static friction.
The normal force is equal to the mass times acceleration due to gravity, but in the opposite direction (negative of the force of gravity).
Since the problem tells us that the force due to friction is 
, we can plug these values into our original equation to solve for the coefficient of friction.
The coefficient of friction has no units.
The equation for the force of friction is
The normal force is equal to the mass times acceleration due to gravity, but in the opposite direction (negative of the force of gravity).
Since the problem tells us that the force due to friction is
The coefficient of friction has no units.
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