All AP Physics 1 Resources
Example Questions
Example Question #21 : Work, Energy, And Power
Object A does work on Object B. The sum of change in potential energy and change in kinetic energy is __________ for object A and __________ for object B.
negative . . . negative
positive . . . negative
positive . . . positive
negative . . . positive
positive . . . negative
Recall that work can be defined as the change in energy of a system. For most systems, we deal with two types of energy: potential and kinetic energy; therefore, work is defined as follows.
Where is work, is change in kinetic energy, and is change in potential energy. Work can be negative or positive based on the type of work being performed. If work is done by an on object (such as object A) then it will have a positive value for work. However, if work is being done on the object (such as object B) then the object will have a negative value for work. We already defined work as being the sum of change in potential energy and change in kinetic energy; therefore, this sum will be positive for object A (positive work) and negative for object B (negative work). Note that both object A and object B will have the same, absolute value for work.
Example Question #21 : Work, Energy, And Power
During takeoff, a rocket goes from at of elevation to at of elevation.
Estimate the amount of work done by the rocket engine. Neglect any mass change due to burning of fuel.
Initially there is only potential energy, and in the final state there is both kinetic and potential.
Plug in all values and solve:
Example Question #23 : Work, Energy, And Power
Determine the work done by pushing a 1kg object up a incline that is long. Assume no friction or air resistance is acting. .
Since the only force acting is gravity, we can determine work by determining the potential energy change.
Since potential energy is given by:
, where is mass, is the gravitational constant, and is height above the earth, we can determine work done by determining:
We can assume
To determine , we determine the height above the ground that the object is moved. We do this by doing:
Therefore,
Example Question #21 : 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).
The rod is initially at rest in its horizontal position. How much work would it take to rotate the rod clockwise until it is vertical, at rest, and mass A is at the top?
Neglect air resistance and internal frictional forces. Ignore the mass of the rod itself.
None of these
We can use the expression for conservation of energy:
Since the rod is both initially and finally at rest, we can removed both kinetic energies. Also, if we assume point p is at a height of 0, we can removed initial potential energy, leaving us with:
Plugging in the expression for potential energy and expanding for both masses:
Since the rod is vertical, we know that mass A is half a rod's length above our reference height, and mass B is half a rod's length below it. Thus we get:
Factoring to clean up our expression:
We know all of our variables, so time to plug and chug:
Example Question #22 : Work, Energy, And Power
A box is being pushed along a frictionless surface by a force, , directed at an angle of below the horizontal. If the block covers a horizontal distance of , how much work, , was done on the box?
The box is moving horizontally across the surface, which means that only the horizontal component of the applied force will do work on the box. The horizontal component of the force is given by
. Therefore, to calculate the work done by the applied force, we will use the standard definition of work, given as
Example Question #24 : Work, Energy, And Power
A box slides down a inclined plane. If , calculate the magnitude of the work done by friction on the box, . Your answer should only have significant figures.
In general,
In order to calculate the work done by a force, we need to find the angle between the force and the displacement through which the object moves. Since the displacement is directed down the incline, then this means the friction force acts up the incline. This means . We must remember the frictional force is represented as
Where is the normal force and is the coefficient of kinetic friction. On an inclined plane, we can show that . Therefore, we can finally write
Plugging in everything, and noting the magnitude is the absolute value of a quantity, we can plug in for our answer:
Example Question #25 : Work, Energy, And Power
A block slides up an incline at a initial speed . The block slides a distance of until coming to a stop. Calculate the work done by the normal force, .
Because the normal force is always perpendicular to the displacement , the normal force does no work on the block, so
This can also be shown mathematically by plugging in zero for theta in the equation for normal force:
Example Question #23 : Work, Energy, And Power
A ball is initially compressed against a spring on a frictionless horizontal table. The ball is the released, and is shot to the right by the spring. Calculate the work done by the spring on the ball, .
Can not be determined.
From the Work-energy theorem,
.
From conservation of energy, we see that all of the potential energy contained in the spring will be transferred into kinetic energy of the ball, shown as
Noting the initial kinetic energy of the ball is , we can see that the work done by the spring will be equal to the initial potential energy of the spring.
Example Question #21 : Work, Energy, And Power
An 500kg elevator is at rest. If it is raised 50 meters and returns to rest, how much total work was done on the elevator?
This can be a tricky question. You need to rely on the work-energy theorem, which states:
Since the elevator is at rest at both the beginning and end, the net work is 0; there is no net change in energy, and therefore no work.
This theorem can be confusing to some since it completely negates potential energy. However, let's think about the situation presented in the problem. A force is required to raise the elevator, meaning that energy is put into the system. However, since it comes back to rest, all of the energy that was put in has been removed by the force of gravity, resulting in a net of zero work.
Example Question #26 : Work, Energy, And Power
A rocket of mass is motionless at the origin. The rocket then fires. At location , it is traveling at .
Determine the work done on the rocket
None of these
Determining magnitude of final velocity:
Plugging in values:
Using definition of kinetic energy:
Plugging in values:
Since there was no initial kinetic energy:
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