May
23

Password to access protected areas of this website

To access any protected areas of this website with physics content please look at the right-hand side of the graphic at the top of the page.  To access copyrighted UIL materials please contact me for your password.

Why have I done this?

In an effort to continue to make my content freely available to my students, parents, colleagues and other educators, but to keep “web-bots” from stealing content from this domain, I have embedded the password information into the graphic at the top of the webpage.  The password should open all publicly accessible portions of this site and you should only have to input it once and your browser will store it (unless you have extremely tight security settings on your web browser).  If you have any problems accessing course content or if the password does not appear to work, please contact me as soon as possible.

Oct
20

AP-1: Vectors and Forces 2, #21 & 22 plus worksheet problems

Newton’s 2nd Law problems from class:  Basic Newtons 2nd problems–KEY
#21: A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads X N.
    (a) As the elevator moves up, the scale reading increases to Y N, then decreases back to X N. Find the acceleration of the elevator.
(b) As the elevator approaches the 74th floor, the scale reading drops to 782 N. What is the acceleration of the elevator?
This problem is actually pretty similar to the rocket problem that we have discussed twice.
First make a free-body diagram.  It is the same for each case: the weight down (mg) and the normal for (N) up.
Then, the “scale reading” in the problem is the weight (mg).  Divide by the acel of gravity to find the mass.  You will need this.
part a)
{F=ma
From the freebody diagram:
N – mg = ma
Normal force is the new scale reading as the elevator rises.  Solve for acceleration.  Should be a positive number indicating the elevator is going upwards.  The reason the scale reading increases is because the floor pushes into your legs.  This makes you feel heavier.
part b)
{F=ma
N – mg = ma
Same as before, right?  Exactly! Because the forces are still the same forces as before, just with new values.  Put your new scale reading into it for N.  Solve for a.  You get a negative acceleration, right?  That’s because the elevator is going down.  You feel light because the floor is dropping out from underneath you.


#22:  A sled of mass X kg is pulled along flat, snow-covered ground. The static friction coefficient is 0.30, and the kinetic friction coefficient is 0.10.

(a) What does the sled weigh?
(b) What force will be needed to start the sled moving?
(c) What force is needed to keep the sled moving at a constant velocity?
(d) Once moving, what total force must be applied to the sled to accelerate it at Ym/s2?

a)  I’m not answering this question for you.

b) Draw a free body diagram.  It should have Your applied force (however much you are pulling…you don’t know this yet) in one direction and your friction force in the other direction.

{F=ma
Fapplied – Ff = ma

Let a = zero.  You only need to apply just enough force to break it free from friction, not make it accelerate. This is an example of the system being in equilibrium in the horizontal direction.  This is very similar math to the “shoes” lab that we did.

Ff=u-static * N, we talked about N in class, but in case you have forgotten it is equal to the weight, which is m*g.  There, you made me answer part a) after all.

So, Fapplied = Ff

c) Same as b, except now you are moving.  Since velocity is constant, a = zero.  Fapplied = Ff, but now Ff is based on kinetic friction.  You need to recalculate it using the coef. of kinetic friction.

d) So now we need the sled to accelerate.
{F=ma
Fapplied -Ff = ma
Fapplied is what you are solving for.
Ff is the force of friction (moving).  This will be the same as the answer to part c, since moving friction was being overcome to keep it at constant velocity.
m = mass
a= your a given in the problem.

Why is your answer to part d) bigger?  Simple: You are having to pull much harder to not only overcome friction and move at a constant speed, but also now to accelerate and speed up.

Problems similar to and expanding on what started at the end of class today.  We will work through some of these tomorrow.

2 body tension without and with friction

2 body tension-pushing #2 from HW

2 body tension–#4 from HW

2+ body tension, like last problem on WebHW

Oct
17

AP-1: Newton’s 2nd Law practice

AP-1–Newtons 2nd practice (force)

Oct
06

AP-1: Solutions to vector addition practice from class

PreAP–KEY–vectors practice (update)

Oct
05

AP-1: Linear Motion test results

AP-1 students:

Your test scores are posted in Skyward.  If you don’t see a grade, be sure to read the note below.  The grade that you see reflects the most favorable, individual score for you.  I took the number of points that you scored on the 2nd part of the test (the cumulative, long part) and divided by just the points on that portion of the exam.  I also took your total points earned on both parts and divided by the overall total.  Whichever percentage was higher is your score that went into skyward.  Sorry it took a while, as you can imagine this required some time to do for all of you.  Many, but not all, students benefited from the overall total, while some students really took advantage of the opportunity to re-learn concepts after the first portion of the exam and did very well on the latter part.  I am thrilled to report that 2 students went from 75% on the first part of the test to a perfect score (100%) on the second part.

Overall there were:

  • 5 perfect scores
  • 16 A’s (17.6%)
  • 28 B’s (30.8%)
  • 64% of students earned 70+
  • Overall average: 76
  • A-Day average: 75
  • B-day average: 77

If you do not have a test score in skyward:

You need to take all or part of your exam.  This must happen before the end of this week, preferably before Thursday.

OR

You did poorly on your exam and you need to see me in tutorial during one of the following times:

  • Monday, after school
  • Tuesday, before or after school (Tuesday PM will be exclusively for you!)

These are the only times that I will be available to work with you on this exam.  Please make time.

Oct
01

AP-C: Force and acceleration analysis

Problems similar to and expanding on what we did in class today.

2 body tension without and with friction

2 body tension-pushing #2 from HW

2 body tension–#4 from HW

2+ body tension, like last problem on WebHW

AP examples

AP-C Sample force problem solutions

Sep
29

AP Physics: Tension Knots lab example

Data is based on system 1, 2011-2013 school year (last couple years) so the final answer is different than the masses this year.  Also the angles will be somewhat different just due to the new set-up.

Tension knots system 1

Tension knots system 3 2012

 

Sep
26

AP-1: Test review materials

In addition to completing your HW and the practice problems from class, here are any documents that I generate for you this weekend, starting with the “quiz” that I didn’t give you in class today.  3rd period you didn’t receive this document on paper (sorry), 4th you got your copy, and B-day you will get yours on Tuesday.  The answer key is on the second page.

AP-1–Linear Motion problems with graphing KEY

Sep
25

AP-1: Linear motion 2 HW helps

#18 & 19:

As discussed in some periods, webassign may score you incorrectly when you try to find the slopes and values of the graphs as indicated in the questions.  Eyeball it to the best of your ability and make sure you know HOW to answer the questions, but I will ensure that you receive credit for items like that that webassign marks incorrect that you actually have right.

#5:

This problem is a common issue for students because (if you read carefully) you will notice that you have two distinct motions represented here.  The knee-jerk approach to this problem is to list out your variables and try to stuff numbers into an equation.  This one is not that easy.  You actually have two separate physics problems here: one is constant velocity motion, one is accelerated (slowing down) motion.

linmot2item5

Your final answer to the question must add the distance the car travels during both time periods, so you first need to find those distances.

Constant Velocity part
If an object is moving at a constant speed, only 1 of our kinematics equations describes its motion: the defn. of velocity–V=(delta)x/t
Based on the information provided in the problem this should be simple to solve for the distance traveled during the reaction time.

Accelerated part
Since this is not constant velocity, the equation we used above goes out the window.  You CANNOT use it because it does not account for the acceleration of the car.  What this does mean, however, is that you can use any of the other 3 equations, provided you have the requisite information.  Here’s what we know:

x = ?          Vo = same speed as first part, since you have not yet started braking              V = 0 (zero, you have come to a stop)

a = given in problem          t = unknown

There are several ways to approach this solution.  You can go directly to the displacement (x) if you select the “right” euqation, but you could also find the time first, then use that information to find the displacement.  It makes no difference.  One important point of caution:  Your acceleration opposes the velocity, so you must make sure it is _________?____________

If you answered “negative” you get a cookie.
Don’t forget to add the 2 distances from the 2 parts of the problem when you finish.

 #7:
I encourage you to approach this question as two separate motions (in some ways similar to number 5, above).  While it can be done as an entire trip (up and then down) it may be easier for you to think about it in two parts.  Use the information provided to find the height to which the ball goes up–don’t forget to account for the fact that gravitational acceleration is down.  Then use the total height to find the impact speed when it falls all the way back to the ground.

Now, on the way down something interesting happens with your problem solving.  If you work this from the max height to the ground then your

displacement is ____?____
velocity is ____?____
acceleration is ____?____

Since everything is negative (you did answer negative or downward for all of those items, right?), you might say that concerns about direction cancel out…they really sort of do mathematically.  In any event you  can solve this using positive numbers provided you recognize that your final answer is downward.  This is important because of the way the question is asked in the problem.  your final answer for your

velocity is ____?____
speed is ____?____

and since the question asks “how fast” your answer should probably be speed.

 

Number 8 is your challenge problem for this assignment.

This is the one problem I will not help you complete fully, but I will give you the following hint:  You need to draw a picture and set up two separate equations for the displacement of the truck (constant velocity) and the displacement of the car (slowing down so need the equation with acceleration).  Write one of the displacements in terms of the other (i.e., 500 – x).  Both of your equations will be missing a displacement and time term.  Solve for time (when they meet the same amount of time must have elapsed).  You will have to either graph it or use the quadratic formula.  Use time to solve for displacement.

Last hint:  Because the car is slowing down fairly rapidly, it will travel the lesser distance.  This may help you narrow down your answer.

If you miss it, don’t worry about the points.  I always factor some participation points into your grades that will offset these type of challenge problems :)

Good luck.

HW #’s 33 & 34 (and notes about 35)

33. A helicopter is rising at X m/s when a bag is dropped from it.

(a) After Y s, what is the bag’s velocity?

(b) How far has the bag fallen?

(c) How far below the helicopter is the bag?

34. A helicopter descends at A m/s when a bag is dropped from it.

(a) After B s, what is the bag’s velocity?

(b) How far has the bag fallen?

(c) How far below the helicopter is the bag?

OK, so these are basically the same problem.  In each problem you know the same information at the beginning.  Since everybody has different numbers I have given variables in the problems here.  For each problem you know:

a = 9.81 m/s^2 down, t and Vo.  In 23 you will want to make your initial velocity positive while the acceleration of gravity is negative.  This means the bag keeps rising for a bit as a result of its inertia but eventually changes direction and falls. In 24 you will want to make everything positive since everything (displacement, velocity, acceleration) is in the downwards direction.

Use the definition of acceleration (a=(delta)V/t) to find the final velocity for both questions part a).  Remember that delta = final – initial.  Your sign changes are very important in this problem!  Your answer may be positive or negative depending on the initial velocity YOU have been given by webassign.  Postive would mean it is still going ___________, but negative means it is going _________.

On part b) I suggest using x=Vot + .5at^2.  In 23 make sure you put in a positive initial velocity and a negative acceleration.  Since x s the displacement this positive and negative discrepancy will take care of the up/down issue and just give you the position of the bag (i.e., “how far it has fallen”) in your result.  If the bag is on its way down (see note from part a) and has fallen past the starting point you will get a negative answer.  The question asked “how far.” This means it wants the absolute value.  In 24 this point is moot since everything is going downward already.  What you get is what you are looking for.

Part c)  These are really your challenge problems on this assignment.  I will give you a big hint, but I am not going to explain completely.  You have two motions here: The helicopter moving at a constant speed (given in your problem), and the falling bag, accelerating due to gravity.  You already know how far the bag has fallen relative to where it started (answer b).  You must find how far the helicopter has gone either up or down (in #23 or 24 respectively) using constant velocity and either add to or subtract from your value from b.  Which one you do depends on the direction of motion.  Draw a before and after picture for each if you are having a hard time seeing it.  Good luck.

For question 35, the scenario is the same as 34, but since the helo is descending the initial velocity of the bag will be downward (negative) along with the acceleration of gravity and the displacement (all negative).  Like part of #7, above, you can drop these negative signs (or keep them as you wish) since all aspects of your motion are down, in the same direction.  Again, take care with part a to ensure that your answer is reflective of direction (down or negative) since the problem specifically asks for “velocity.”

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