# High School Physics : Understanding Conservation of Energy

## Example Questions

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### Example Question #1 : Energy And Work

A skier starts at the top of a hill with  of energy. Assuming energy is conserved, what is her final kinetic energy?

Insufficient information to solve.

Explanation:

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 .

### Example Question #1 : Understanding Conservation Of Energy

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 skier is at the bottom of one hill, but will go back up another

The skier miscalculated her energies

The skier is not a very good skier

The skier must have paused somewhere during her descent

Work must have been done

Work must have been done

Explanation:

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.

### Example Question #1 : Energy

A rock is dropped in freefall from some initial height . Which of the following describes its final velocity right before it hits the ground?

Explanation:

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.

### Example Question #4 : Energy And Work

book falls off the top of a  bookshelf. What is its kinetic energy right before it hits the ground?

Explanation:

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.

### Example Question #1 : Energy

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?

Explanation:

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:

### Example Question #1 : Understanding Conservation Of Energy

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?

We need to know the mass of the skier to solve

Explanation:

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.

### Example Question #7 : Energy And Work

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 .

### Example Question #8 : Energy And Work

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.

Explanation:

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 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.

### Example Question #1 : Energy

Computational

A ball is dropped from  above the ground. Assuming gravity is , what is its final velocity?

Explanation:

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.

### Example Question #10 : Energy And Work

A sled is initially given a push up a frictionless  incline.  It reaches a maximum vertical height of .  What was its initial speed?

Explanation:

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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