All College Physics Resources
Example Questions
Example Question #1 : Mechanics
A toy car is set up on a frictionless track containing a downward sloping ramp and a vertically oriented loop. Assume the ramp is tall. The car starts at the top of the ramp at rest.
What additional piece of information is necessary to calculate the maximum height of the loop if the car is to complete the loop and continue out the other side?
The value of
The mass of the car
None
The exact shape of the loop
The distance between the end of the ramp and entrance to the loop
None
This is an example of conservation of energy. The car starts at the top of the ramp, at height . It has no velocity at this time since it is starting from a rest. Therefore its total energy is where is the mass of the car and is the value of gravitational acceleration.
At the bottom of the loop, all of the potential energy will have been converted into kinetic energy.
As the car traverses the loop and rises above the ground, kinetic energy will be converted back into potential energy. The shape of the loop does not matter in this case -- only the vertical distance between the ground and the car.
In the tallest possible loop, all kinetic energy at the bottom is converted to potential energy at the top. This is the maximum height the car can reach -- there is no additional energy left to continue climbing a taller loop. Therefore, the potential energy at the top of the tallest loop we can build is equal to the kinetic energy at the bottom of the loop. But we have already noted that the kinetic energy at the bottom of the loop is equal to the potential energy at the top of the ramp.
Therefore, we set . We see that and cancel, and we are left with . In other words, the tallest loop you can build is equal to the height of whatever ramp you select. In this example, the tallest loop we can build is . We do not need to know the specific values of or .
Example Question #2 : Mechanics
An elevator is designed to hold of cargo. The designers want the elevator to be able to go from the ground floor to the top of a tall building in . What is the minimum amount of power that must be delivered to the motor at the top of the shaft? Assume no friction and that the elevator itself has a negligible weight.
Power is the rate of energy transfer. To raise a object , a total of or ( is required. To find the power in Watts (), we divide the total energy required by the time over which the energy must be transferred:
Example Question #2 : Work And Energy
How far can a person jump while running at and a vertical velocity of ?
We know that:
and we are looking for the maximum height (vertical displacement) this person can obtain, so we aren't concerned with .
We can apply the conservation of energy:
Masses cancel, so
Solve for :
(rounded to simplify our calculations)
so let's plug in what we know
. This is our final answer.
Example Question #1 : Conservation Of Momentum And Energy
If a object has a kinetic energy of right after it is launched in the air, and it has KE at its max height, what is its max height?
Let's first write down the information we are given:
In order to solve this problem we must apply the conservation of energy, which states since no friction.
This means that as the project reaches its max height energy is converted from Kinetic Energy (energy of motion) to potential gravitational energy (based off of height).
We can subtract from to get the at its max height
=
so we can solve for the height
where
therefore
Example Question #2 : Conservation Of Momentum And Energy
If an object has a kinetic energy of right after it is launched in the air, and it has KE at its max height of , what is the object's mass?
Let's first write down the information we are given:
In order to solve this problem we must apply the conservation of energy, which states since no friction.
This means that as the project reaches its max height energy is converted from Kinetic energy (energy of motion) to potential gravitational energy (based off of height).
We can subtract from to get the at its max height
=
so we can solve for the mass
where
therefore