alzamanan+m


 * 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. **


 * (x-xo)= 190 **
 * v= 5.08 m/s ( transfer from m/min to m/s " 350/60") **
 * a= 1.22 m/s^2 **


 * i). How far does the cab move while accelerating to the maximum speed starting from rest? **

// from the number we found in the equstion i will use this equation [ V^2= Vo^2 + 2a (X-Xo) ] // // (5.08)^2= 2(1.22)^2 (X-Xo) then // // (X-Xo)= (5.08)^2/ 2(1.22)^2= __10.58 (M)__ //


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

// in this part to found T i will use ( V= Vo + a* t ) // // v= 5.08 m/s // // Vo=00 // // a= 1.22 m/s^2 // // t =?? // // 5.08 m/s = 1.22 m/s^2 *t // __** t = 4.16 s **__


 * 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-xo)= 50 m // // Vo= 00 // // a=9.81m/s^2 // // t=?? // // i will use this equation // // (x-xo)=vo*t+ 0.5 a*t^2 // // 50 m = 4.9 m/s^2 * t^2 // // t^2 = 50 m / 4.9 m/s^2 // __//t1 = 3.19 s//__

// t2 = 2.19s ( t1- 1 = 2.19s) // // and // // Vo=?? //
 * ii). What is the initial speed of the second stone? **

// same the first part's equation //

// (x-xo)=vo*t+ 0.5 a*t^2 // // 50 m=Vo * 2.19s + 0.5*(9.81m/s^2) (2.19s)^2 // __// Vo = 12.1 m/s //__ Professor's Note : Good effort !! 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**

__//we ahve to use those tow egoustion//__ __//A+b = 6.0 iˆ + 1.0 jˆ//__ __//and//__ __//A-B= - 4.0 iˆ + 7.0 jˆ//__

__// get the final answer for each, then add them together and solving for B //__ __//sqrt (Ax+Ay)^2 that for A//__ __//and//__ __//sgrt(Bx+By)^2 for B//__ __//add the i^ for X and the j^ for Y//__
 * // ii. Find the magnitudes of A and B //**

iii. **//What is the angle between A and B//** __//we have to use the dot porduct to found the angle//__ **Professor's Note** : Good work, explain part iii bit more or visit my page for more details


 * 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 of all, we have to figure out the X and Y components

As V0x= V0 cos (θ)

V0y=V0 sin(θ)


 * Then we have to found the T
 * Then we have to use this equation to found it.

(x-x­o) = Vox*t + 1/2a*t²


 * Then, found the height by using this

(y-yo­) = Voy*t + 1/2(ay)*t^2

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?

Part (1)

We have (y-yo) and ( V0y ).

First we have to find the time by using this equation:

(y-yo) = (Voy)*t+ ½ a *t^2

Then by this equation (x-xo) = (Vox)*t+ ½ a*t^2 we can solve for Vox.

To find the initial velocity we have to set this equation and solve for it

v = sqrt ((Vox)^2 + (Voy)^2 )

Part (2)

Find the time by using this equation ( y-yo) = (Voy)*t+ ½ a*t^2

After that we can find Vox by using this equation (x-xo) = (Vox)*t+ ½ a*t^2 -

Professor : No new work is found !


 * Week 06 -- EXAM WEEK --- No wikis**
 * 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 (b) with a speed 8 m/s while decelerating at a rate of 2.4 m/s²

(A)we know that the acceleration is 0 because the speed is constant. now we have to use this equation (t=mg+ma) and the ma=0, t=65

(B)also we know the acceleration is decreasing the it will be -2.4 and mg=65 for solving this problem we have to use this equation (t=mg+ma) then we will end with some numbers for m. then plug it in same equation to have the final answer.


 * __ 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 planeb). 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?

Part (A) Given fs = Ms.Fn, M static = 0.25 and Fn = m*g cos theta. sum of F = 0

to solve for F also find the magnitude F + fs - mg.sin(theta) = 0 Part(B) Given

fs = Ms.Fn, M static = 0.25 and Fn = m*g cos theta.

to solve for F and the min force

F - fs - mg.sin(theta) = 0

Part(C)

we have the equation fk = Mk.Fn, M kinetic = 0.15 and Fn = m*g cos theta then F - fk - mg.sin(theta) = 0 constant is velocity then A = 0

now solve it for F and find the value of F (force) has to move up the plane at constant velocity

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,

Draw a picture.

A. Find the work done by the cord's force on the block

W= F·dcosƟNeed to find FF=maTcos(0) + mgcos(180)=maT= mg+maW=(mg+m(g/4)·d **answer in joulesB. Find the work done by the gravitational force on the blockW=F·dcosƟF=mgW=mgdcos(0)W=mgdD. Find the speed of the blockV2=Vo2 +2a(Δd)Vo is zeroV= Sqrt(2(g/4)d) C. Find the kinetic energy of the blockKE =1/2mv2 Use the v from part dKE=1/2m(sqrt(2(g/4)d)

// Part(a) //

// To find the work done by the cord's force on the block we have to find part B //

// wg= m * (9.81 m/s^2)* d * cos (0) from Part (B) //

// now combine the eqoution we got from part (B) with Newton’s second law Fn= m*a //

// then we got //

// wt=td cos(Ɵ)=m(a+g)d cos(Ɵ) //

// Now substitute downward acceleration in the equation and we know the block is moving downward so the angle will change to 180degrees now use this equation //

// Wt=(g/4)*(m)*(9.81m/s^2)*(d)*cos(180) //

// Part(B) //

// By using this equation (Wg=mgd cosƟ the block move downward then we get positive direction and the angle is zero) to find the work done by the gravitational force on the block //

// Then we got //

// wg= m * (9.81 m/s^2)* d * cos (0) //

// Part(C) and (D) //

// We have to use this equation V^2=Vo^2+2a(X-Xo) to find velocity //

// And we have //

// Vo=0, (X-Xo)=d , a=(-g/4) //

// So we get this eqoution //

// V= //

// Using this eqoution //

// Kf=Ki+W=.5*m*v^2+W //

// .5*m* //// +W //

// Now find the network during the fall by using W=Wg+Wt //


Week 10 (April 16th, 2012)

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

Hint: first draw a diagram. find the velocity of the ball, just before it hits the block.

Steps :

I went through some answers, and did not find the correct final answer or the steps anywhere.

(Just found out Steve G has the right approach - please visit his page as well !)**

Here are my steps:

1. Draw a diagram

2. Use the conservation of energy equations, accounting the lowest poison (where the block is) as the zero potential level

--> initial velocity of the ball = 0 m/s thus KE (initial) = 0 --> Initial potential energy = ugh where h is the length of the cord --> final KE = 1/2 mv² --> final potential energy = 0

using the above information and placing them in, (KE + PE )i = (KE + PE)f

you can find the final velocity of the ball, v, before it hits the block

3. Now we are considering the conservation of momentum

m1v1i + m2v2i = m1v1f + m2v2f (1) we know, m1, m2, v1i and v2i we don't know, v1f or v2f --> to find two un-knowns we need two equations

4. We consider elastic collision

(Total KE )before = (Total KE)after

(1/2m1v1i² +1/2m2v2i²) = (1/2m1v1f² +1/2m2v2f²) (2)

again we know m1, m2, v1i and v2i