Tim+Madore

__**Week 2 (Jan 30th 2012 )**__


 * 01. A certain cab has a total run of 190 m and a maximum speed of 305 m/min. It accelerates from rest and back to rest at 1.22 m/s².**
 * i). How far does the cab move while accelerating to the maximum speed starting from rest?**

First 305 meters/minute needs to be converted to meters/second which equals 5.083 meters/second

solving for displacement plug into the equation v^2=v0^2+2(a)(displacement) (5.083)^2=0+2(1.22)(x) solve for x
 * x=10.59 meters **
 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest?**

since we have the displacement for an acceleration to full speed, and the cab decelerates at the same rate, we know that the time for each is equal. therefore solve for time during one of those accelerations solve for t x=v0*t+.5(a)(t)^2

10.59=0+.5(1.22)(t)^2 --- wrong displacement -PROFESSOR t=4.166 seconds 

190-(10.59*2)= 168.82 meters v=d/t 5.083=168.82/t t=33.21 seconds

2*(4.166)+33.21=**41.542 seconds**

a=9.81m/s^2 t=? y=50m
 * 02. A stone is dropped into a river from a bridge 50 m above the water. Another stone is thrown vertically down 1.00 s after the first stone is dropped. The two stones st rike the water surface at the same time.**
 * i). How long does it take for the first stone to reach the water surface?**

x=v0*t+.5(a)(t)^2 50=0+.5(9.81)(t)^2
 * t=3.192 seconds **

time=3.192 seconds- 1 second because thrown 1 second after the first y=50m a=9.81m/s^2 v0=?
 * ii). What is the initial speed of the second stone?**

x=v0*t+.5(a)(t)^2 50=v0*(3.192-1)+.5(9.81)(3.192-1)^2
 * v0=12.0585 m/s **

Professor's Note : Good effort - MORE explanations please !! Do not include numerical values in the final answer -- check my page

__**Week three (06th of February, 2012)**__

1. If **B** is added to **A** the result is 6.0 iˆ + 1.0 jˆ. If **B** is subtracted from **A**, the result is - 4.0 iˆ + 7.0 jˆ.

i. Find **A** and **B** A+B=6i^+1j^ A-B=-4i^+7

(A+B)+(A-B)== A (A+B)+(A-B)== B

ii. Find the magnitudes of **A** and **B**

The equation for magnitude of a vector simply uses the pythagorean theorem. It can be changed around so that

|**A**|=sqrt((Ax)^2+(Ay)^2) and |**B**|=sqrt((Bx)^2+(Ay)^2)

iii. What is the angle between **A** and **B** use dot product formulas; one without angle and one with and make them equal to eachother **A·B=**Ax*Bx+Ay*By and

**A·B=**|**A**|*|**B**|*cos(ø) - couldn't find the theta symbol

Ax*Bx+Ay*By=|**A**|*|**B**|*cos(ø) all values are known except for theta(phi) solve for theta ø=arccos((Ax*By+Ay*By)/|**A**|*|**B**|)

__**Week four (13 th February, 2012)**__

1. A 0.5 kg rock is projected from the edge of the top of a building with an initial velocity of 10.5 m/s at an angle 54º above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 15.3 m from the base of the building. Assume: The ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s². What is the height of the building?

mass=.5kg --negligible, mass isn't taken account for in any of the equations vi=10.5 m/s theta=54 deg g=9.8 deltaX=15.3 m deltaY=?

use the deltaX equation and solve for time deltaX=Vi*cos(theta)(time)

plug (t) into deltaY equation to solve for deltaY deltaY=Vi*sin(theta)(time)-.5(g)(time)^2

deltaY is the height of the building

__**Week 05 ( 20th February, 2012)**__

A pelican flying along a horizontal path drops a fish from a height of 5.4 m. The fish travels 8.0 m horizontally before it hits the water below. The acceleration of gravity is 9.81 m/s2. a) What was the pelican’s initial speed? b) If the pelican was traveling at the same speed but was only 2.7 m above the water, how far would the fish travel horizontally before hitting the water below?

draw diagram with vectors and positions

make two table of known values deltaY=5.4 m deltaX=8 m  Gy=9.81 m/s^2 Gx=0 Vix=? t=?

A) plug known values into equations to solve for time  deltaY=viy(t)+.5(g)(t)^2  5.4=0(t)+.5(9.81)(t)^2  t=

plug time into equation again but solve for vix

deltax=vix(t)+.5(g)(t)^2 8=vix(t)+.5(0)(t)^2 vix=

B) first solve for time at 2.7 meter freefall  deltaY=viy(t)+.5(g)(t)^2  2.7=0(t)+.5(9.81)(t)^2  t=

then use that time to solve for new deltaX DeltaX=vix(t)+.5(0)(t)^2 DeltaX==vix(t) DeltaX=

Tim, I visited the page, Will do it again : try to post the work by 9 PM, and don't worry, if you don't see the comments, I have allocated the extra credits for all your hard work above !


 * Week 07 ( 05th of March, 2012)**

01 . An object is hung from a spring balance attached to the celing of an elevator cab. The balance reads 65 N, when the cab is standing still. What is the reading when the cab is moving upward,

(a) with a constant speed of 8 m/s and If the speed of the cab is constant there is no acceleration no acceleration means that there will be no change to the tension from the balance to the object because the equation f=ma a=0 and that means that an additional force also equals 0 the scale will stil read 65 newtons

(b) with a speed 8 m/s while decelerating at a rate of 2.4 m/s² since a=-2.4 m/s^2, this time there is an additional force acting on the object plug into the equation sumforces=ma where a is equal to -2.4 Tension-mg=m(-2.4) solve for mas using the force the balance was reading when cab was standing still w=mg plug back into tension equation and solve for the tension because that is the force that the balance would be reading when the cab is decelerating

I will try to post earlier next time

__** Week 08 (March, 12th, 2012) **__

01. A loaded sled weighing 85 N rests on a plane inclined at an angle of 20⁰ with the horizontal. The coefficents of static and kinetic friction between the inclined plane and the sled are 0.25 and 0.15 respectively.

a). What is the least magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane

b). What is the minimum force required to start the sled moving upward?

c). What value of F (force) is required to move the sled up the plane at constant velocity?

must find maximum static friction in the opposite direction of F when the least force equals the maximum static friction, there will be no acceleration of the sled (not moving) so sum of forces including µFs equal to m(0) ""not moving" solve for F magnitude solve for FN in the same manor and plug back into the equation to find the minimum force to start moving

__**Week 9 (April 26th, 2012)**__

1) A cord is used to vertically lower an initially stationary block of mass M at a constant downward acceleration of g/4. When the block has fallen a distance d i) Find the work done by the cord's force on the block W=F*d*cos(theta) F=ma Tcos(90)+mgcos(270)=ma W=T*d*cos(theta) ii) Find the work done by the gravitational force on the block W=F*dcos(theta)  F=mg  W=mg*d*cos(270)

iii) Find the kinetic energy of the block KE=1/2mv^2  >>>>> insert V from part iv KE=1/2m(sqrt(2*(9.81/2)*d)^2 iv) Find the speed of the block a=9.81/4 m/s/s  V^2=Vi^2+2*a*d  V=sqrt(2*(9.81/4)*d)>>>>> for part iii