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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:

• Marble Down Incline
• First Encounter with Basic Motion
• Average Rates, Average Rates of Change of Position and Velocity
• Graphs of s, v and a vs. t, including interpretation

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. 

• How much work did I do?

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

• a = Fnet / m = 20 N / (25 kg)

= .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:

• How much work did friction do on the cart?
• Assuming a constant frictional force, what is the force of friction?

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:

• Initially the cart had KE equal to .5 m v^2 = .5 ( 25 kg) ( 2 m/s)^2 = 50 kg m^2 / s^2 = 50 Joules.
• At the end the KE was 0, since vf is zero.
• So `dKE = 0 Joules - 50 Joules = -50 Joules.
• Therefore `dWnet = `dKE = -50 Joules.
• `dWnet is the work done by the net force.

The only force acting in the direction of motion is the frictional force.

• So we know that the frictional force did - 50 Joules of work over a displacement of 2 meters.

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

• Fave = `dW / `ds = -50 Joules / (2 meters) =

-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.