Jeff+P.

=__ **WEEK 2** __=

a) First convert 305m/min to m/s. (305m/min)/(60s) = 5.08m/s b) Using the equation [V f^2=(Vi)^2+2a( ΔX )], sub in the values (plug & chug) c) 5.08^2=0+2(1.22)(Xf-Xi) => 25.80=2.44( Δ X) => ΔX =10.576m
 * 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? **

Answer = 10.58m

a) Because the acceleration and deceleration is uniform, the scenario is parabolic and thus we can use the formula [Vf =Vi+at] b) (Plug & Chug) 5.08m/s=0+1.22m/s^2(t) ---> wrong use of initial and final velocities and the sign of "a" c)5.08/1.22=t => t=4.16s d) Because the acceleration and deceleration are the same and the equation is parabolic, multiply the answer by 2 to get the proper answer.
 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest? **

Answer = 8.32s


 * 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 strike the water surface at the same time. **
 * i). How long does it take for the first stone to reach the water surface? **
 * i). How long does it take for the first stone to reach the water surface? **

a) Use the equation [ Δ x=Vit+(1/2)at^2] b) (Plug & Chug) 50m=0t+(1/2)(9.8m/s^2)t^2 c) Simplify. 50=4.9t^2 => 50/4.9=t^2 => 10.20=t^2 => t=sqrt(10.20) d) t=3.19s

Answer = 3.19s

a) Use the equation [Δ x=Vit+(1/2)at^2] b) (Plug & Chug) 50m=Vi(3.19s)+(4.9m/s^2)(3.19-1.00s)^2 c) Simplify. 50=(3.19-1.00s)*Vi+23.50 => 26.5=2.19s*Vi d) Vi=12.10m/s
 * ii). What is the initial speed of the second stone? **

Answer = 12.10m/s


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

=__** WEEK 3 **__=


 * 01. 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) First, we must combine the two resultant vectors to determine what is A and what is B. We get (A+B)+(A-B)=2A

(6.0i^+-4.0i^)+(1.0j^+7.0J^)=2A
 * b) Now, we just solve for vector A.

A=(2.0i^+8.0j^)/2 A=1.0i^+4.0j^
 * c) Now, we must find what vector B is now that we have found vector A.

(A+B)=6.0i^+1.0j^ => B=(6.0i^+1.0j^)-(1.0i^+4.0j^) B=5.0i^-3.0j^

Vector A=__1.0i^+4.0j^__ Vector B=__5.0i^-3.0j^__


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


 * a) To find the magnitudes of both A and B, we can use Pythagorean theorem (make sure to replace units at the end).

A(magnitude)=sqrt(i^2+j^2) A=sqrt(1.0^2+4.0^2) A=sqrt(17)≈__4.12m__

B(magnitude)=sqrt(i^2+j^2) B=sqrt(5.0^2-3.0^2) B=sqrt(16)=__4.0m__


 * iii. What is the angle between A and B**


 * a) To find the angle between vectors A and B, you can use the dot product rule. Since we have the magnitudes of A and B, this is an easy task.

The formula for the dot product rule is as follows: A.B**=**|A||B|cosø. Also, A.B=AxBx+AyBy, so we can use this to our advantage. Using the two equations we can set AxBx+AyBy=|A||B|cos ø. Now all we do is insert the values and come up with an answer.

(1.0i **.**5.0i^)+(4.0j^**.-**3.0j^)=|4.12||4.0|*cosø Solve for ø => (5)+(-12)=|16.48|*cosø => (-7)/16.48=cosø => arccos(-7/16.48)=ø ø =115.13°

Therefore, the angle between vectors A and B is __115.13°__

**Professor's Note: great job, post the work by 9 PM**

**Visited the page today - will visit again** = = =__ **WEEK 4** __=


 * 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?**


 * First, we must find the X and Y components of the initial velocity. We use the formulas given:

V0y = 10.5sin(54) = 8.49 m/s V0x = 10.5cos(54) = 6.17 m/s


 * Then, we find how long the ball is in the air using a horizontal equation. Since the acceleration in the horizontal direction is zero, we can use:

Xf-Xi = V0x * t 15.3 = 6.17 * t t = 2.47 s


 * Then, we can plugin the time collected in the horizontal direction for the vertical direction. Note that in the following equation, g is the gravitational constant, 9.8 m/s^2. All other values have already been computed.

Yf-Yi = (V0y * t) - (1/2 *g * t^2) Yf-Yi = (8.49 * 2.47) - (1/2 * -9.8 * 2.47^2) Yf-Yi = 50.86 m


 * Normally, we would utilize the other number given in the original problem, that the rock weighs 0.5kg, but since we have all unknowns, there is no need to include mass/weight.

Therefore, the height of the building is __50.86m__

= = =__ **WEEK 5** __=


 * 01. 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?**


 * First we must use the equation Xf-Xi=Vx*t. We do not have Vx or t, so we must look at the vertical equation. Yf-Yi=Vy*t-(1/2)(9.81)t^2. The only unknown in this equation is t.

-5.4=0+(-4.95)(t^2) t=5.4/4.95 t=1.09s

> 8=Vx*(1.09) Vx (initial/final x-velocity)=7.34 m/s
 * Then, we can solve for the unknown in the first equation Xf-Xi=Vx*t


 * This makes the initial speed of the pelican 7.34 m/s in the positive horizontal direction.


 * If the pelican was traveling at the same speed but only 2.7m above the water, we can use the same equations to figure out the range.

Yf-Yi=Vy*t-(1/2)(9.81)t^2 -2.7=0+(-4.95)(t^2) t=2.7/4.95 t=0.545s

Xf-Xi=Vx*t Xf-Xi=(7.34)(.545)


 * Long story short, if the vertical displacement is halved, the horizontal displacement will similarly be halved.

Xf-Xi (or the range)=4.0m

= __**WEEK 7**__ =


 * 01 . An object is hung from a spring balance attached to the ceiling 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 **
 * (b) with a speed 8 m/s while decelerating at a rate of 2.4 m/s² **


 * First we can formulate the equations for force in the y-direction, since in the x-direction there is no acceleration and therefore F=0

First, let's define parameters. Ft = force of tension of the spring balance/reading on the spring balance.

Σ Fy=Ftsin90+mgsin270=ma Σ Fy=Ft-mg=ma Ft-mg=ma Ft=ma+mg

Also, for reference we can determine the mass of the object with Ft=mg Ft/g=m m=6.63kg

Because the acceleration is zero for part a, Ft=mg, which means the reading will continue to be __65N__


 * However, when the cart is decelerating....

Ft=(-2.4)m+mg Ft=(-2.4)(6.63)+(6.63)(9.8) Ft=__49N__

Professor's Note : Jeff, you are doing a wonderful JOB. keep up the good work !!

=__ **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,**

__**A. Find the work done by the cord's force on the block**__ W= F·dcosƟ Need to find F F=ma Tcos(0) + mgcos(180)=ma T= mg+ma W=(mg+m(g/4)·d **answer in joules** __B. Find the work done by the gravitational force on the block__ __D. Find the speed of the block__ __C. Find the kinetic energy of the block**__ KE =1/2 mv2  Use the v from part d  KE=1/2 m(sqrt(2(g/4)d)
 * W=F·dcosƟ **
 * F=mg **
 * W=mgdcos(0) **
 * W=mgd **
 * V2=Vo2 +2a(Δd) **
 * Vo is zero **
 * V= Sqrt(2(g/4)d) **

= __WEEK 10__ =


 * A steel ball of mass 0.5 kg is fastened to a cord that is 70 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of it's path the ball strikes, a 2.5 kg steel block initially at rest on a friction-less surface. The collision is elastic. **


 * Find, **
 * a) the speed of the ball **
 * b) the speed of the block, both just after the collision **

First, we need to figure out the speed of the steel ball at the bottom of it's pendulum's path. If the kinetic energy is 0 when released and if the potential energy is 0 at the bottom of the path, we can use conservation of energy to solve for the speed.

mgh=(1/2)mv^2

Masses cancel, so we are left with:

gh=(1/2)V1^2 V1=(2gh)^0.5,

Where h is the vertical displacement, or, the length of the cord which is 0.70m. Now solve for V1 (Velocity of the ball directly before collision)... Once we have that we can use the conservation of momentum with knowledge the collision is elastic to find the speed of the block after the collision, V2.

M1V1+ M2V2 = M1V3 + M2V4 (0.5)sqrt(2gh) + 0 = (0.5)V3 + (2.5)V4

Solve for V3. V3 = ((0.5)sqrt(2gh) - (2.5)V4)/0.5

Now utilize conservation of kinetic energy:

(1/2)M1V1^2+ (1/2)M2V2^2 = (1/2)M1V3^2 + (1/2)M2V4^2 (1/2)M1V1^2+ 0 = (1/2)M1V3^2 + (1/2)M2V4^2 M1V1^2 + 0 = M1V3^2 + M2V4^2 ((0.5)sqrt(2gh)^2 - M1V3^2 = M2V4^2 V4^2 = (((0.5)sqrt(2gh)^2 - (0.5)*[((0.5)sqrt(2gh)^2 - (2.5)V4)/0.5]^2) / M2

Solve for V4, which is the speed of the block after collision. Then input that value back into the kinetic energy equation to solve for V3 which is the speed of the ball after collision.