Handouts to look at:

Any of these have good information but the 4th talks about Atwood systems and tension forces.




Use www.physicsclassroom.com

Physics Classroom Handout that we did before:
http://www.physicsclassroom.com/getattachment/curriculum/force2D/forces.pdf

Video:

http://www.learnapphysics.com/apphysicsb/kinematics.php

Elevator Problems:

http://www.batesville.k12.in.us/physics/phynet/mechanics/newton2/ElevatorProblem.html


Atwood Problems:



2 Box Problem

the two box question on the review work (19)

Forces Review:



32989011.jpg
external image 971425_644604392227242_282670096_n.jpg

Labs:

Lab 1

Lab 2

Lab 3

Notes:

Note: the problem where you have a cart going up an incline and another cart is released with velocity at the same time the first cart stops and both carts reach the bottom of the incline at the same time. (two part question-find the distance up the incline the first cart goes and the initial speed of the second cart)
First of all you need to calculate the acceleration component acting along the incline. This is your gravitational field.

Now you can calculate how long for the first sled to go up and stop. (This will be the same time for it to go back down again.) And also how far up it goes.

Knowing that you know how long the second sled has to reach the bottom and hence what speed it must be released at.


Block on wedge block with horizontal force:
Forces1.png

Discussion


DustinS.101
For problems that involve two blocks one on a table with friction the other suspended from a pulley, connected to the first block and hanging off the table. When using the Fnet=ma formula the m is the mass of block one plus the mass of block two. The acceleration and net force for block one is the same as the acceleration and net force for block two.

Using Newton’s second law, the accelera-
tion of the wedge of mass 18 kg and block of
mass 5 kg is
aAB =
F
mA + mB
(1)
Consider the free body diagram for block B
m g sin 
N
mg
aAB
The condition that block B does not move
with respect to wedge A implies that its ac-
celeration down the wedge with respect to the
wedge is zero. Applying Newton’s second law
for B in the direction parallel and perpendic-
ular to the wedge yields
XF? : N − mB g cos  = mB aAB sin (1)
XFk : mB g sin  = mB aAB cos  (2)
Using Eq. 1 to solve for the normal force