Physics 30 2012 Period 3

• To start off the class we looked at Jeren's blog about Newton's third law of motion.


• We finished c) on example one of the notes finding out that we would add 22N to 121N to get 143N so it will overcome the force that is used by toboggan 2.


• We than got assigned practice questions on page 149, numbers 1, 2, 3a. Once done that we could finish assignment 6 that we got a few days ago. Once done those two assignments we could work on the review on page 155, numbers 8, 9, 11, 13, 17-20.

Page 149 Question #1

Two girls, one of mass 40 kg and the other of mass 60kg, are standing side by side in the middle of a frozen pond. One pushes the other with a force of 360N for 0.10s. The ice is essentially frictionless.

(a) What is each girl's acceleration?

Mg1-40kg Mg2-60kg Fa-360N T-0.10s

F=ma F/m=a 360N/40kg=a =9m/s^2 360N/60kg=-6m/s^2

(b) What velocity will each girl acquire in the 0.20s that the force is acting?

Vf=Vi+at =0+ 9m/s^2(0.10s)^2 = 0.9m/s

Vf=0+(-6m/s^2)(0.10s) = -0.60m/s

(c) How far will each girl move during the same time period?

d=Vi(t)+1/2(a)(t)^2 +1/2(9m/s^2)(0.10)^2 = 4.5 x10 ^-2 m

D=1/2(-6m/s^2)(0.10)^2 = -3.0 x 10 ^-2 m

2) Two crates of masss 12.0kg and 20.0kg, respectively, are pushed across a smooth floor together, the 20kg crate in front of the 12kg crate. Their acceleration is 1.75m/s^2. Calculate each of the following.

(a) the force applied to push the crates.

M1=12.0kg M2=20.0kg a=1.75m/s^2

F=(32.0kg)(1.75m/s^2) = 56.0N

(b) the action-reaction forces between the two crates.

F=(20.0kg)(1.75m/s^2) = 35.0N

Dana is next