All High School Physics Resources
Example Questions
Example Question #1 : Work Kinetic Energy Theorem
What is the work applied to a car when it is pushed up along a incline at a constant velocity.. Assume the coefficient of friction between the car and the ramp is .
Known
Unknown
The work done on the object is equal to the net force done on the object multiplied by the displacement through which the force acts.
Therefore the first thing to do in this problem is to determine the amount of force needed to push the car up the incline. Since the car is moving at a constant velocity, all forces must be balanced. The force in the direction of motion is the force applied which is doing work on the object. The forces opposing motion are the force of friction and the x-component of the gravitational force.
To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction. Using trigonometric functions we get that
We know that the force of gravity is equal to mg
The force of friction is directly related to μ (the coefficient of friction) times the normal force. In this case the normal force is equal to the y component of the force of gravity.
Therefore
We know that our force applied must balance both of these forces.
We can now substitute this equation into our work each to determine the work applied to the car.
We can now substitute in our variables and solve for the missing piece.
Example Question #2 : Work Kinetic Energy Theorem
At an accident science on a level road, investigators measure a car’s skid mark to be long. The accident occurred on a rainy day, the coefficient of friction was estimated to be . What was the speed of the car when the driver slammed on and locked the brakes?
Knowns
Unknown
We know that the force of friction is what caused the skid marks and is the cause of the work done on car to slow it to a stop. We also know that the work is equal to the change in kinetic energy.
Work is equal to the force times the displacement over which the force acted.
Therefore
The force in this case is the force of friction.
Since the car is on a level road the normal force is equal and opposite the force of gravity.
The force of gravity is equal to the mass times the acceleration due to gravity
Therefore
We can substitute this in for our work-kinetic energy equation.
Since the final velocity is we can cancel this out of the problem.
Since there is a negative and a mass in both terms we can cancel these out of the equation
Now we can substitute in our values and solve.
Example Question #1 : Work Kinetic Energy Theorem
A cat knocks a toy mouse across the floor with of force. If the cat did of work, how far did the mouse travel?
The relationship between work, force, and distance is:
We are given the force on the toy and the work done. Using these values, we can find the distance. Note that the mass is not relevant for this question.
Example Question #1 : Work Kinetic Energy Theorem
A rocket has a net propulsion force of . Over a short period of time, the rocket speeds up uniformly from an initial velocity of to a final velocity of . Assume that the mass of the rocket remains constant even though it is burning fuel and that the net force is along the direction of travel. What is the net work done on the rocket in ?
Use the data given to calculate the kinetic energy of the rocket at the two different velocities. Then find the amount of work done using the following equation:
Kinetic energy of the rocket at the two velocities:
The change in the kinetic energy at the two velocities:
Example Question #1 : Work Kinetic Energy Theorem
A bodybuilder lifts a weight upward , and then returns it to its original position, all with a constant speed. What is the net work done on the weight?
Work is a vector quantity equal to the change in energy in a given direction. Our net work will be equal to product of force and displacement.
We can write this equation in terms of Newton's second law to incorporate our given values.
During its motion the weight travels a distance of 0.5m upward, and then 0.5m downward. Its total distance will be 1.0m, however, we are looking for its displacement. The initial height and final height are equal, making the displacement equal to zero.
The total work will be zero because there was no net displacement.
Example Question #1 : Work Kinetic Energy Theorem
Which of the following is an example of work?
A skier skis down a frictionless hill with no loss in kinetic energy.
A student types numerous essays for class.
A weightlifter lifts a heavy barbell above his head.
A car drives along the road at a constant speed.
A man sits in a chair.
A weightlifter lifts a heavy barbell above his head.
Work is a force times a distance or .
Only one of these has a force times a distance: the weightlifter lifting a weight above his head.
The car moving at a constant velocity has no acceleration and therefore no force.
Typing essays does not require a force times a distance (though it sure feels like work!).
If the skier has no loss in energy, then no work was done, as work is also the measurement of the change in energy.
The man sitting in a chair has a constant velocity of zero, no acceleration, and therefore no force.
Example Question #1 : Work Kinetic Energy Theorem
A arrow is fired from a bow whose string exerts an average force of on the arrow over a distance of . What is the speed of the arrow as it leaves the bow?
Known
Unknown
We can use the work kinetic energy theorem to solve this problem. The work done to the object causes a change in kinetic energy.
Work is equal to the force times the displacement over which the force acted.
Therefore
Plug in our variables and solve
Example Question #5 : Work Kinetic Energy Theorem
A book falls off the top of a bookshelf. How much work is required to put the book back on the top of the bookshelf, assuming it is lifted with a constant velocity?
Work is a force times a distance:
We know the distance that the book needs to travel, but we need to solve for the lifting force required to move it.
There are two forces acting upon the book: the lifting force and gravity. Since the book is moving with a constant velocity, that means the net force will be zero. Mathematically, that would look like this:
We can expand the right side of the equation using Newton's second law:
Use the given mass and value of gravity to solve for the lifting force.
Now that we have the force and the distance, we can solve for the work to lift the book.
This problem can also be solved using energy. Work is equal to the change in potential energy:
While on the ground, the book has zero potential energy. Once back on the shelf, the energy is equal to . The work is thus also equal to .
Example Question #6 : Work Kinetic Energy Theorem
Rodrigo pushes a sofa down a long hallway. If the couch moves with a constant velocity, how much work does Rodrigo do?
We need to know the mass of the sofa to solve
We need to know the coefficient of friction of the sofa on the floor to solve
Work is the product of a net force over a given distance.
In order for there to be work, there must be a net force applied and the object must have a non-zero displacement. In this question we are given the displacement, but we have to solve for the force.
The question only tells us that the couch moves with constant velocity. This tells us that acceleration is zero. If acceleration is zero, then force must also be zero according to Newton's second law.
If the force is zero, then the work is also zero.
Example Question #59 : Energy
How high will a stone go if thrown straight up by someone who does of work on it? Ignore air resistance.
Known
Unknown
When something does work on an object it changes the total energy of the object. In this case, the work done by the person converts to kinetic energy as the stone is launched. This kinetic energy is then turned into gravitational potential energy when the stone is at the highest point of the peak.
Work done = Kinetic Energy when the stone leaves the hand = Gravitational Potential Energy at the peak
We can set the work done to the gravitational potential energy
Plug in our known values and solve.
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