Energy and Work
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Physics › Energy and Work
A spring with hangs vertically next to a ruler. The end of the spring is next to the
mark on the ruler. If a
mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?
Explanation
Known
Unknown
First let us consider the force acting on the spring. Since the spring is hanging vertically, the only force acting on the spring is the force of gravity from the mass that has been added to the spring.
We can substitute in our variables and find the gravitational force.
According to Hooke’s Law the spring constant of a spring is directly proportional to the force acting on it and inversely proportional to the amount of stretch.
We can use the provided spring constant and the force acting on the spring to determine the amount of stretch.
Find the minimum initial height of the roller coaster if the roller coaster is to complete the
diameter loop.
12.1 m
37.5m
48.3m
40.2m
50.8m
Explanation
First, we need to determine how fast the roller coaster must be going at the top of the loop to continue in a circular motion. At the top of the loop, the only force acting on the car is gravity. Therefore the gravitational force must be the cause of the centripetal motion.
We know that the force of gravity is
And the centripetal force equation is
We can set these two equations equal to each other.
Since mass is on both sides of the equation we can cancel it out.
We can rearrange and solve this equation for the velocity.
We can now use the conservation of energy to determine the initial height of the roller coaster. We know at the top of the roller coaster, there is only . At the top of the loop of the coaster there is both
and
.
Since mass is each factor, we can cancel it out.
The height of the coaster must start at .
A box is dropped
. How much work was done on the box?
Explanation
The formula for work is , work equals force times distance.
In this case, there is only one force acting upon the object: the force due to gravity. Plug in our given information for the distance to solve for the work done by gravity.
Remember, since the object will be moving downward, the distance should be negative.
The work done is positive because the distance and the force act in the same direction.
A box is dropped
. How much work was done on the box?
Explanation
The formula for work is , work equals force times distance.
In this case, there is only one force acting upon the object: the force due to gravity. Plug in our given information for the distance to solve for the work done by gravity.
Remember, since the object will be moving downward, the distance should be negative.
The work done is positive because the distance and the force act in the same direction.
A spring with hangs vertically next to a ruler. The end of the spring is next to the
mark on the ruler. If a
mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?
Explanation
Known
Unknown
First let us consider the force acting on the spring. Since the spring is hanging vertically, the only force acting on the spring is the force of gravity from the mass that has been added to the spring.
We can substitute in our variables and find the gravitational force.
According to Hooke’s Law the spring constant of a spring is directly proportional to the force acting on it and inversely proportional to the amount of stretch.
We can use the provided spring constant and the force acting on the spring to determine the amount of stretch.
A spring with hangs vertically next to a ruler. The end of the spring is next to the
mark on the ruler. If a
mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?
Explanation
Known
Unknown
First let us consider the force acting on the spring. Since the spring is hanging vertically, the only force acting on the spring is the force of gravity from the mass that has been added to the spring.
We can substitute in our variables and find the gravitational force.
According to Hooke’s Law the spring constant of a spring is directly proportional to the force acting on it and inversely proportional to the amount of stretch.
We can use the provided spring constant and the force acting on the spring to determine the amount of stretch.
Find the minimum initial height of the roller coaster if the roller coaster is to complete the
diameter loop.
12.1 m
37.5m
48.3m
40.2m
50.8m
Explanation
First, we need to determine how fast the roller coaster must be going at the top of the loop to continue in a circular motion. At the top of the loop, the only force acting on the car is gravity. Therefore the gravitational force must be the cause of the centripetal motion.
We know that the force of gravity is
And the centripetal force equation is
We can set these two equations equal to each other.
Since mass is on both sides of the equation we can cancel it out.
We can rearrange and solve this equation for the velocity.
We can now use the conservation of energy to determine the initial height of the roller coaster. We know at the top of the roller coaster, there is only . At the top of the loop of the coaster there is both
and
.
Since mass is each factor, we can cancel it out.
The height of the coaster must start at .
A box is dropped
. How much work was done on the box?
Explanation
The formula for work is , work equals force times distance.
In this case, there is only one force acting upon the object: the force due to gravity. Plug in our given information for the distance to solve for the work done by gravity.
Remember, since the object will be moving downward, the distance should be negative.
The work done is positive because the distance and the force act in the same direction.
Find the minimum initial height of the roller coaster if the roller coaster is to complete the
diameter loop.
12.1 m
37.5m
48.3m
40.2m
50.8m
Explanation
First, we need to determine how fast the roller coaster must be going at the top of the loop to continue in a circular motion. At the top of the loop, the only force acting on the car is gravity. Therefore the gravitational force must be the cause of the centripetal motion.
We know that the force of gravity is
And the centripetal force equation is
We can set these two equations equal to each other.
Since mass is on both sides of the equation we can cancel it out.
We can rearrange and solve this equation for the velocity.
We can now use the conservation of energy to determine the initial height of the roller coaster. We know at the top of the roller coaster, there is only . At the top of the loop of the coaster there is both
and
.
Since mass is each factor, we can cancel it out.
The height of the coaster must start at .
A crate, starting from rest, is pulled across a floor with a constant horizontal force of
. For the first
the floor is frictionless and for the next
the coefficient of friction is
. What is the final speed of the crate?
Explanation
For this we can consider the work-kinetic energy theorem. As work is done on the object, its kinetic energy is changing. In this case we have two different situations to consider. In the first we must consider the horizontal force acting on the box alone. In the second we must consider the horizontal force being resisted by a frictional force.
Let’s begin with the horizontal force acting alone.
Work is equal to the force times the displacement of the object.
In the first section the only force is and the displacement is
.
We can use the work kinetic energy theorem to solve for the change in kinetic energy during this first section
Since the initial velocity is zero the equation becomes
We can now plug in our values
This is the velocity of the box after the first . Now it is time to analyze the motion of the box when it has both friction and the applied force.
Newton’s 2nd law says that the net force is equal to the sum of the forces involved.
We need to find the friction force.
The normal force in this case is equal to the force of gravity
We can now determine the work on the box through the next .
Like we did before we can now find the change of kinetic energy. This time we will use the final kinetic energy from the first part as the initial kinetic energy of the second part.
Therefore the box will have a final velocity of .