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A rock is dropped in freefall from some initial height . Which of the following describes its final velocity right before it hits the ground?
For this problem, we must use the law of conservation of energy.
Since the initial velocity is zero, there is no initial potential energy. Since the final height is zero, there is no final potential energy. This means that the final kinetic energy equals the initial potential energy.
The mass can be canceled from both sides.
Now we need to isolate the velocity by multiplying both sides by two, and taking the square root.
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A skier starts at the top of a hill with of potential energy. At the bottom of the hill, she has only
of kinetic energy. Assuming that at the top of the hill she has only potential energy and at the bottom she has only kinetic energy, what can we conclude?
The work-energy theorem states that work is equal to change in energy, or .
Total mechanical energy is the sum of potential and kinetic energies:
In this case, she starts with and ends up with
. Since there was a change of
, that means at some point during the system,
of work was done by the skier.
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A book falls off the top of a
bookshelf. What is its kinetic energy right before it hits the ground?
Remember the law of conservation of energy: the total energy at the beginning equals the total energy at the end. In this case, we have only potential energy at the beginning and only kinetic energy at the end. (The initial velocity is zero, and the final height is zero).
If we can find the potential energy, we can find the kinetic energy. The formula for potential energy is .
Using our given values for the mass, height, and gravity, we can solve using multiplication. Note that the height becomes negative because the book is traveling in the downward direction.
The kinetic energy will also equal , due to conservation of energy.
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A skier starts at the top of a hill with of energy. Assuming energy is conserved, what is her final kinetic energy?
If energy is conserved, then the total energy at the beginning equals the total energy at the end.
Since we have ONLY potential energy at the beginning and ONLY kinetic energy at the end, .
Therefore, since our , our kinetic energy will also equal
.
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A man stands on a tall ladder of height . He leans over a little too far and falls off the ladder. If his mass is
, what is his kinetic energy right before he hits the ground?
For this problem, use the law of conservation of energy. This states that the total energy before the fall will equal the total energy after the fall. The initial kinetic energy will be zero, and the final potential energy will be zero; thus, the initial non-zero potential energy will be equal to the final non-zero kinetic energy.
We can use the energy equations to define these equal energies:
The energies are equal, so we can say:
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A skier waits at the top of a hill. He then skis down the slope at an angle of
above horizontal. What will his velocity be at the bottom of the hill?
To solve this problem, use the law of conservation of energy. The skier initially starts at rest; all of his initial energy will be potential energy. At the bottom of the hill, the potential energy will be zero and all of the final energy will be kinetic energy. We can set these two values equal to one another based on the conservation of energy principle.
Expand this equation to include the formulas for potential and kinetic energy.
Notice that the mass cancels out from both sides. This allows us to calculate without knowing the mass of the skier.
Plug in our given values for the height of the slope and acceleration due to gravity. Since potential energy is a state function (independent of the path) the slope of the hill is irrelevant.
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A projectile is fired at an upward angle of from the top of a
cliff with a speed of
. What will be its speed when it strikes the ground below? (Use conservation of energy to solve this problem)
Knowns:
Unknowns:
The easiest way to solve this problem is by analyzing the energy of the object at different points during its path. The first point to analyze is the moment it is launched from the top of the cliff. Assuming that the ground is our zero point in reference, the projectile has both Gravitational Potential Energy and Kinetic Energy when it is launched.
At the end when the projectile lands on the ground, it no longer has gravitational potential energy. It now is all
We know that based on the law of conservation of energy, the total energy at the beginning must equal the total energy at the end. Therefore
Our equations for each are as follows
We can substitute these equations in.
Notice that mass is in every term so we can cancel it out. That is why mass does not matter for this problem.
Substitute in our values and solve for the final velocity.
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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?
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 .
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Mike jumps off a bridge with a bungee cord (a heavy stretchable cord) tied around his ankle). He falls for before the bungee cord begins to stretch. Mike's mass is
and we assume the cord obeys Hooke’s law. The
constant is
. If we neglect air resistance, what is the distance below the bridge Mike’s foot will be before coming to a stop. Ignore the mass of the cord and treat Mike as a particle.
We must consider several points during Mike’s jump off of the bridge. The first point is when he is at the top of the bridge when he is about to jump. The second point is the below the bridge, just when the bungee cord would begin to stretch. The third is the point at the bottom of the cord when it is fully stretched out.
To start let, us consider the first two points, when he jumps off the bridge and when he reaches below the bridge. For this first consideration, I will assume that our zero point of reference is
below the bridge.
At the top of the bridge, Mike has gravitational potential energy. later, all of this potential energy has been converted to kinetic energy. According to the law of conservation of energy we can set these two things equal to each other.
Since mass is in both sides of the equation it can be cancelled out to leave us with
We can now solve for the final velocity, just before the cord stretches.
Now let us consider two new points, the point at which the cord starts to stretch, and the point at the bottom when the entire cord is stretched out. We will consider the lowest point as our zero point of reference in this case.
At the top, Mike has kinetic energy and gravitational potential energy as he is moving and above our reference point. At the bottom all of this energy has converted to elastic potential energy. According to the law of conservation of energy we can set these two things equal to each other.
The cord is going to stretch the same distance that Mike starts above the ground so we can exchange our x value for h so that everything is in similar terms.
We can now put in our values and start to solve for h. We will use our velocity from the first part as the velocity that Mike has.
We are left with a quadratic equation. So we will need to get everything over to one side and use our quadratic formula to solve this problem.
The quadratic formula is
The two answer we get for this is and
. The reasonable answer is
. This is the distance the cord will stretch.
To find the total distance below the bridge we will need to add the amount that the cord stretched to the it took to fall before the cord stretched.
Mike will stop below the bridge.
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Computational
A ball is dropped from above the ground. Assuming gravity is
, what is its final velocity?
We can use potential energy to solve. Remember, your height and your gravity need to have the same sign, as they are moving in the same direction (downward). Either make them both negative, or use an absolute value.
Using conservation of energy, we know that . This tells us that the potential energy at the top of the hill is all converted to kinetic energy at the bottom of the hill. We can substitute the equations for potential energy and kinetic energy.
The masses cancel out.
Plug in the values, and solve for the velocity.
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A sled is initially given a push up a frictionless incline. It reaches a maximum vertical height of
. What was its initial speed?
We can use conservation of energy to consider the energy at the top of the incline and the bottom of the incline. At the bottom of the incline the sled has some velocity. At the top of the incline the sled has gravitational potential energy. According to the law of conservation of energy these two values must be equal.
The mass cancels out of the equation.
The angle does not matter in this case because it is a frictionless surface and all energy is conserved.
The initial velocity of the sled is .
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A child descends a slide
high and reaches the bottom with a speed of
. How much thermal energy due to friction was generated in this process?
We can use conservation of energy to solve this problem. Let us consider the types of energy at the beginning and the end. At the beginning the child has gravitational potential energy at the top of the slide. When the child reaches the bottom, the child has both kinetic energy and thermal energy as some energy was converted to heat because of the friction on the slide.
The law of conservation of energy states that we can set the energy at the beginning equal to the energy at the end.
This difference between the
at the top and the
at the bottom is the energy lost to friction.
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Conceptual
A man stands on a tall ladder of height . He leans over a little too far and falls off the ladder. If his mass is
, what is his velocity right before he hits the ground?
For this problem, use the law of conservation of energy. This states that the total energy before the fall will equal the total energy after the fall. The initial kinetic energy will be zero, and the final potential energy will be zero; thus, the initial non-zero potential energy will be equal to the final non-zero kinetic energy.
From there, expand the equation to include the individual formulas for potential and kinetic energy calculation.
Notice that the mass will cancel out from both sides.
Now we can solve for the final velocity in terms of the initial height.
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A runner arrives at the bottom of a hill. He runs up the hill with a constant acceleration until he reaches the top, then runs at a steady pace along the top of the hill. When are the kinetic and potential energies of this man at their greatest?
To answer this question, we can address each type of energy separately. There is no conservation of energy in this problem; kinetic energy is not converted to potential energy as the man runs up the hill. Instead, he is accelerating, indicating an outside force that disallows conservation of energy.
First, we will find the maximum potential energy using the equation:
The man's mass and the acceleration of gravity will remain constant. The only changing variable is height. When the height is greatest, the potential energy will be the greatest. We can conclude that the potential energy will thus be greatest at the top of the hill.
Now we will look at the equation for kinetic energy:
The man's mass will remain constant, and the only changing variable will be the velocity. We are told that the man accelerates as he runs up the hill, indicating that his velocity is increasing. This tells us that he will reach a maximum velocity when he reaches the top of the hill, at which point he maintains a steady velocity along the top of the hill. Since kinetic energy is at a maximum when velocity is at a maximum, we can conclude that kinetic energy is greatest at the top of the hill.
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Two balls are thrown off a building with the same speed. One is thrown straight up, and one at a angle. Which statement is true if air resistance is ignored?
We can use conservation of energy to analyze this problem. When both balls are thrown off the building, they both have the same initial gravitational potential energy. Additionally, since both balls are thrown with the same magnitude of the speed, they both have the same kinetic energy. Energy is a scalar quantity and therefore does not have a direction. Therefore it is independent of the path taken.
At the bottom, both balls will have converted all their potential energy and kinetic energy that they started with to kinetic energy at the bottom of the building. Since both started with the same amount of total energy at the beginning, their kinetic energy at the end will also be the same. Since both balls are assumed to have the same mass, their magnitude of their velocity (speed) will be the same as well. However, their velocity will be in different directions because of the way that the ball was launched.
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Find the minimum initial height of the roller coaster if the roller coaster is to complete the
diameter loop.
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 .
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