All AP Physics 1 Resources
Example Questions
Example Question #221 : Work, Energy, And Power
Consider the following system:
Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at the midpoint between the masses and is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. is the angle at which the rod makes with the horizontal at any given time ( in the figure).
This system is initially held so that the rod is in a horizontal position. At , the system is released and the rod is allowed to rotate. How much work is done by gravity by the time mass A reaches the lowest point of rotation?
Neglect air resistance. Assume the mass of the rod is zero.
We can use the work-energy theorem to solve this problem
Now let's use the equation for conservation of energy:
Rearranging for change in kinetic energy, we get:
Substituting this into the work-energy theorem, we get:
We no longer care about the velocity of either mass. We're simply concerned about the change in potential energy.
Using the following equation:
We can expand the later two terms:
If we assume that the bottom of the circle has a height of 0, we can say that the final potential energy of mass A = 0. Therefore, we get:
Substituting these into the work-energy-theorem, we get:
Factoring out g to make things neater:
With our last assumption, we can say that the initial height of both masses
and the final height of mass B
Now that we have values for everything, it's time to plug and chug:
Example Question #21 : Power
Consider the following scenario:
A sledder of mass is at the stop of a sledding hill at height with a slope of angle .
If a sledder of mass is traveling down the hill that has a slope of at a constant velocity of . How much power is kinetic friction doing on the sledder? Neglect air resistance and other frictional forces.
There are multiple paths to solve this problem, but we will go with the most straight forward. There are two forces acting on the sledder in the direction of their motion: kinetic friction and a component of the gravitational force. The problem statement tells us that the sledder is traveling a constant velocity, so we know that these forces are equal and opposite. Therefore we can calculate the component of gravitational force and set that equal to the frictional force. We can then use that and the sledder's velocity to calculate power. So to begin, the force of gravity:
However, this is a vertical force and we need the component that is in the direction of the sledder. Therefore, we will multiply by sin:
If you're wondering why sin was used, think about the situation practically. As the angle grows, the hill gets steeper, and there will be more gravitational force in the direction of acceleration. Sine is the function that gets larger as the angle gets larger.
We can now set this equal to the frictional force:
Plugging in values for this:
Now to get the power that this force uses, we can multiply it be the velocity that it is applied at. If this is confusing, take it in steps. If we multiple a force by distance, we get energy. If we multiple energy by frequency, we get power. And distance times frequency is simply velocity. Moving on:
Example Question #22 : Power
Calculate the power required from a forklift to move a box vertically through a distance of in a time of at constant velocity.
By definition,
Because the box is moving at constant velocity, we know that the work done by the forklift must be equal to the work done by gravity, written as:
Combining this with the time we would like to raise the box, we can calculate the power,
Example Question #23 : Power
If an truck engine must produce a force in order to maintain a constant speed of up an incline, calculate the power produced by the engine.
Cannot be determined.
Power in general is written as
In this form, the problem is not able to be solved. However, if we think a bit outside of the box, we can massage the expression for power in such a way to solve our problem. Consider
Using our altered definition of power, we can calculate:
Example Question #24 : Power
A locomotive accelerates a train from to in . Estimate the power of the locomotive during this time period.
None of these
All energy is kinetic, and the initial energy is zero, so:
Combining equations:
Converting minutes to seconds and plugging in values:
Example Question #261 : Ap Physics 1
A cyclist maintains an average velocity of on flat surfaces. In order to overcome air resistance and maintain that velocity, the cyclist must be constantly accelerating at . If the cyclist has a mass of , how much power is being generated by the cyclist?
To have power, an object needs to have a force and a velocity:
The acceleration comes from the air resistance that's trying to constantly decelerate the cyclist, and the velocity is the speed the cyclist is maintaining:
Example Question #26 : Power
If three locomotives are pulling a train, how much power does each locomotive need to apply on average during the first second to accelerate the train at from rest?
Using
Using
and
All energy will be kinetic. Combining equations:
Plugging in values:
Example Question #221 : Newtonian Mechanics
During time period , a rocket ship deep in space of mass travels from to . During time period , the rocket fires. During time period , the rocket travels from to .
Time periods , , and took each.
Determine the average power during time period .
None of these
Using
Determining initial kinetic energy:
Combining equations
Converting to and plugging in values:
Determining final kinetic energy:
Combining equations
Converting to and plugging in values:
Plugging in values:
Using
Example Question #221 : Work, Energy, And Power
A hot air balloon with mass has begun its descent and is traveling downward at a rate of . What is the gravitational power loss of the balloon per second?
The expression for gravitational potential energy is:
And power is:
Therefore, we get:
Plugging in our values, we get:
Example Question #28 : Power
Determine the power needed to lift a woman in at a constant velocity.
None of these
The work will be equal to the gravitational energy change:
Using