Chris+Giza


 * 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? **

The first step is to convert the units. ((305m/min)/(60s))=5.08m/s

Next, to find the distance, we use the equation and plug in the variables. v^2=v0^2+2a(x-x0) (5.08m/s)^2=2(1.22m/s^2)(x) (25.81m^2/s^2)/(2.44m/s^2)=x X=10.58m


 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest? **

For this problem we use the equation. V=v0+at Plug the variables into the equation. 5.083m/s=(1.22m/s^2)t (5.083m/s)/(1.22m/s^2)=t t=4.17s Don’t forget to double the result. 4.17s*2=8.35s


 * 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? **

First, we use the equation. (x-x0)=v0t+(1/2)a(t^2) plug in variables 50m=(1/2)(9.8m/s)(t^2) 50m=(4.9m/s)t^2 Square root (50m/4.9m/s)=t t=3.19


 * ii). What is the initial speed of the second stone? **

First, we use the equation. (x-x0)=v0t+(1/2)a(t^2) plug in variables 50m=v0(2.19s)+(1/2)(9.8m/s)(2.19s)^2 50m-23.5=v0(2.19s) 26.5=v0(2.19s) initial velocity v0=12.10m/s

Professor's Note : I must have missed the work last week ! Good job. try to avoid numericals

__ **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 ** From looking at the equation, we can see that the best way to find A and B is to simply use basic addition and subtraction to find the results that work to find the results of B added to A and B subtracted from A.

ii. Find the magnitudes of ** A ** and ** B ** In order to find the magnitudes of A and B, the Pythagorean Theorem is used. Example… Sqrt(i^2+j^2)=magnitude

iii. What is the angle between ** A ** and ** B ** In order to find the angle between A and B, we know that the results of A and B will produce a right triangle. Moreover, because of this triangle, we know that side A and side B is not the hypotenuse (or magnitude) therefore, arc tan can be used in order to find the angle between A and B. Example… Tan(i/j)^-1= angle between A and B

Professor's Note : Good work !! comment 1) on part ii) try not to use the unit vectors, instead use the coefficients Ax and Ay in the square-root comment 2) Explanation on part iii doesn't appear to be correct. Visit my page

visited the page and will do again!

__ **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?

For this problem, we first need to find the time in which the rock was in the air. We do this by using information given about the horizontal... Vo=10.5m/s g=0 X-Xo=15.3m t=?

We plug the numbers into the equation to get the time... X-Xo=(Vo*cos (54))t

Next, we need to find the height of the building by using the equation for the y component... Y-Yo=? Vo=10.5m/s t=2.48 g= -9.8

Plug numbers into the equation for the height of the building... (Y-Yo)= (Vo*sin(54))t-(1/2)*g*t^2

__ **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?

A) In order to find the the initial speed of the pelican, we know the vertical height and the acceleration. We plug what we know into the equation y-yo=(Voy)t+(1/2)a*t^2 to find the t (time). However, Voy is zero since the pelican is at its maximum height. so the equation we have is y-yo=(1/2)a*t^2.

Next, we use the equation x-xo=(Vox)*t+(1/2)a*t^2, where the acceleration is 0 so we use x-xo=(Vox)*t. Now, in order to find the initial speed, plug in the variables.

B) Since the horizontal variable changes, we need to use the Voy that was found in part a. Find the time using, y-yo=(1/2)a*t^2, after plunging in the variables.

Now, using the variables, we can find the distance the fish would travel horizontally by using the equation x-xo=(Vox)*t+(1/2)a*t^2, with the acceleration equal to 0.

__ **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

We know that when the cab is stationary, the acceleration is equal to zero. Also, when an object has aconstant velocity, acceleration is also zero. T-mg=ma a=0 mg=65 using this we can solve for T

(b) with a speed 8 m/s while decelerating at a rate of 2.4 m/s² a=(-2.4) this is negative because of the y direction is decelerationg. we know T-mg=ma with mg=65 we can use this to find the mass by dividing the weight by the gravity.

Professor's Note : No new work posted ! will visit again (I apologize for not being able to post this sooner.)

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.

First thing to do is to draw a force diagram of the problem.

a). What is the least magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane Use the static coefficient in order to find the sled as it starts tomove. fs=Coefficient of static*(W*cos(20))

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

Because F=ma, we know that our equation of forces will be equal to zero because the sled is not moving. F*cos(o)+fs*cos(108)+mg*cos(250)+Fn*cos(90)=0 next, replace all of the cos with sin for the y component. Once equation's are reduced, the minimum of force can be found when set equal to each other.

c). What value of F (force) is required to move the sled up the plane at constant velocity? We know that because velocity is constant, acceleration is 0. This means that the sum of the forces is equal to zero and that the when the object is at equilibrium, there is a constant velocity. Similar to when an object is at rest. Fnet=0

Week 9 (April 6th, 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 ii) Find the work done by the gravitational force on the block iii) Find the kinetic energy of the block iv) Find the speed of the block

i) in order to find the work done by the cord's force on the block, first part 2 must be completed.

ii) In order to find the work done by the gravitational force, we use the equation Wg=mgd cos(theta) in this case, because the block is moving downward, this is the positive direction and the angle of theta is zero. so...Wg=m*(9.81 m/s^2)*d*cos(0)

i) Now that we know the equation form the gravitational force, in order to find cords's force on the block, we need to combine the equation from part 2 with Newton's second law...Fnet=ma. From that equation, we get T-Fg=ma in order to find the tension of the cord on the block.

when we combine the two equations we get...

Wt=Td cos(theta)=m(a+g)d cos(theta)

Next, we need to substitute the downward acceleration into the equation (-g/4) and because we know the block is moving downward, the theta in this case now changes to 180 degrees.

Wt=(g/4)*(m)*(9.81m/s^2)*(d)*cos(180) (note:the answer work we get should be negative in this case)

iii) In order to find the kinetic energy, first the speed of the block needs to be found in step iv.

iv) In order to determine the velocity of the block, we use the equation V^2=Vo^2+2a(X-Xo) and plug in the varables we know. Vo=0 (X-Xo)=d a=(-g/4)

V=square root(0+2(-g/4)(d))

iv) Now that we know have the velocity, we can plug that into our equation for kinetic energy. Kf=Ki+W=(1/2)mv^2+W =(1/2)*m*(square root(0+2(-g/4)(d)))^2+W

Lastly, we need to find the net work done on the cab during the fall, in order to plug that into our equation. In order to do that, we need to add the Wg and the Wt from the previous questions above. W=Wg+Wt W=m*(9.81 m/s^2)*d*cos(0)+((g/4)*(m)*(9.81m/s^2)*(d)*cos(180))

so the final equation we get for the kinetic energy is... Kf=(1/2)*(m*(square root(0+2(-g/4)*(d)))^2)+(m*(9.81 m/s^2)*d*cos(0)+((g/4)*(m)*(9.81m/s^2)*(d)*cos(180)))

reduced we get... Kf=(1/2)*(m*(2(-g/4)*(d)) + (m*(9.81 m/s^2)*d) - ((g/4)*(m)*(9.81m/s^2)*(d))

Week 10 (April 16th, 2012)

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 frictionless surface. The collision is elastic. Find, a) the speed of the ball

In order to find the speed of the ball, we need to set the equations for potential energy equal to kinetic energy, shown below.

mgh=(1/2)mv^2

m is the mass of the ball, g is 9.81m/s^2, and h is the height of the ball in its initial position, but in this case its the same as the length of the cord which needs to be converted to meters.

Using this we can find the velocity of the ball before it hits the block.

b) the speed of the block, both just after the collision

Since we know the speed of the ball before it hits the block, we can find the speed of the block using the equation for conservation of momentum, shown below.

mi1*vi1+mi2*vi2=mf1*vf1+mf2*vf2 m1 is the mass of the ball m2 is the mass of the block v1 is the velocity of the ball v2 is the velocity of the block f means final i mean initial

Using the variables for mass and velocity, we can determine the final velocity of the block because conservation of momentum states that all energy can ony be sifted and not destroyed.