0927
Note for reference: Practice Major Quiz
change default page on each computer to http://vhmthphy.vhcc.edu
Note for reference: Linked Outline
Path to class notes is physics I > assts, scroll to bottom and click on Fall 04.
Click on today's date and look at Synopsis with Multiple Choice Question References
At h:\shares\physics run the Multiple Choice Questions choosing the 201 course. Run the following:
Include comments when something isn't clear or when you think you see an error. If you think there's an error start your comment with the word 'error'.
I just exerted a force of 20 Newtons as I pulled the cart at a low constant velocity through a displacement of 4 meters.
I did 20 Newtons * 4 meters = 80 Joules of work.
A Joule is a Newton * meter. A Newton is a kg m/s^2. So a Joule is a kg m/s^2 * m = kg m^2 / s^2.
So the work I did was 80 kg m^2 / s^2.
The mass of the cart is about 25 kg. If the cart had experienced a net force of 20 N while travling 4 meters, how fast would it have been going at the end, assuming that it started from rest?
We know Fnet, m, `ds and v0. From Fnet and m we can get the acceleration a, which can then be combined with `ds and v0 to get vf.
We find that
= .8 N / kg = .8 ( kg m/s^2 ) / kg = .8 (kg m/s^2) * (1 / kg) = .8 m/s^2.
We can then use the fourth equation of motion and the fact that net force is in the positive direction to conclude that vf = +sqrt( v0^2 - 2 a `ds) = ... = + 2.56 m/s.
If something with mass m starts from rest with a net force Fnet accelerating it thru displacement `ds parallel to the net force, how fast will it goin' at the end?
We know Fnet, m, `ds and v0.
We can find a = Fnet / m.
Then since vf^2 = v0^2 + 2 a `ds we can say that
vf^2 = v0^2 + 2 (Fnet / m) `ds, which we rearrange to get
Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2.
Defining kinetic energy (abbreviated KE) as 1/2 m v^2 and noting that Fnet `ds = `dWnet we get
`dWnet = `d( KE).
This is the work-energy theorem.
I just pushed the cart, giving it an initial velocity of 2 m/s. It coasted to rest over a distance of 2 meters.
Use the work-energy theorem to answer:
The only force acting in the direction of motion is the frictional force (note that the force I exerted to achieve the initial velocity is no longer acting; that force ceased at the instant the initial velocity was achieved). So the net force is the frictional force.
The work-energy theorem tells us that `dWnet = `d(KE), so from the change in KE we can find the work done by the net force:
The only force acting in the direction of motion is the frictional force.
From the work done by a force and its displacement we can find the average force exerted over that displacement.
A force that does -50 Joules of work over a 2 meter displacement exerted an average force of
-25 J / m =
(-25 kg m^2 / s^2) / m =
-25 kg m/s^2 =
-25 Newtons.
Using a spring balance, which reads about 2.8 kg when pulling the cart at a constant velocity across the floor, we find that the frictional force is equal to the force that would be exerted by gravity on a 2.8 kg mass. How does our result compare with our prediction?
Gravity exerts a force of magnitude 2.8 kg * 9.8 m/s^2 = 27 Newtons on a 2.8 kg mass, so the magnitude of our frictional force is about 27 Newtons. This compares well with our result from the work-energy theorem.
Interpret F vs. x graphs, including linear graph for pendulum.