class 050919

What would be the units of each of the following:

• .5 a `dt
• a `ds
• 2 a `dt^2
• 2 `ds `dt

In Friday's experiment some of you had 'blocks' made of 2 dominoes, some 3 and some 4. 

For a given 'slingshot', assuming the same pullback for each:

• Which would you expect to gain the greatest speed upon release?
• Which do you think would slide furthest?
• Which do you think would gain the most energy from the rubber bands?

List each pullback you used to project the blocks, the area corresponding to that pullback, the sliding distance and the number of dominoes in your block:

Sketch your force vs. pullback distance graph for the 'slingshot' you used Friday on the back.  Make it a quick stetch, showing just the shape of the graph and labeling the maximum pullback and maximum force.

more practice with interpretation of graphs

practice problems for Major Quiz

Introductory Problem Sets:  Click on Lectures at top of your homepage, then on Introductory Problem Sets.  (If this doesn't work click on Assts then scroll to Asst 1 and look for Introductory Problem Sets, Set 1, #'s 1-7; click on the link).

Be sure you know how to do everything in Introductory Problem Sets 1 and 2.

In class you had said that the area of the sliding distance will equal the area under the rectangle (that we know how to
calculate). For one of the trials our rubber band had a pull back of -6.5, a force of 0.5, and a sliding distance of 17 cm.
I calculated the area to be 1.625 cm* N. I know the area for the sliding distance would be the same but how do I calculate
it with these numbers.

Excellent question.

1.625 cm * N, rounded off to 1.6 N * cm because your estimate of the area under a curving graph won't be accurate to more
than 2 significant figures, is about right for the work under the first graph, according to your figures.

The area under the rectangle representing frictional force for a sliding distance of 17 cm should therefore be -1.6 N * cm.
The frictional force is nearly constant, so the region would be a rectangle. Its dimensions would be f_friction * 17 cm,
where f_friction stands for the frictional force.

So f_friction * 17 cm = -1.6 N * cm.

Solving for f_friction we get

f_friction = -1.6 N * cm / (17 cm) = -.093 N, approximately.

We'll go over this in class.

The equation is from problem 33 and their are 3 different equations and they want you to tell which equations can work and which one's can not.

The equations are
X= vt^2 + 2at
X= v(initial)t + .5at^2
X= v(initial)t + 2at^2
I don't know what to put in for X when I am sloving the equations using the units from 1-8.

<h3>This isn't one of the equations but here is an example:

If the equation is `ds = v0 `dt + .5 a `ds, which is not one of the equations of motion but at first glance looks OK, you check out the units of each term of the equation.

Using MKS units:

The units of `ds are meters.
The units of v0 `dt are m / sec * sec = meters.
The units of a `ds are m / sec^2 * m = m^2 / sec^2.

Since you can add unlike terms, you can't add meters to m^2 / sec^2, and the right-hand side is not dimensionally consistent.</h3>