Laura+C

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 **
 * Use the conservation of energy to find the velocity of the ball. **
 * this will be v3 **
 * 1/2 m v^2 + mgh = 1/2 m v^2 +mgh **
 * mgh = 1/2 m v^2 **
 * sqrt(2gh)= v3 **
 * Then use the conservation of momentum to solve for the velocity of the block after the collision **
 * MbV1+MxV2= Mbv3+MxV4 **
 * V1+V2= V3+V4 **
 * (V1+V2)-(V3) =V4 **

The end __ -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
 * Draw a picture
 * the coefficient for static friction is used.
 * solve for Fn
 * Fn=85cos(20)
 * F= Fn*(mus)
 * Plug in Fn and the coefficeint and solve

b). What is the minimum force required to start the sled moving upward?
 * use kinetic friction because the object is now moving
 * max=Fkcos(0)+Fncos(90)+mgcos(250)
 * Fx=Fk+mgcos(250)
 * Fx=coefficentof kinetic*85+mgcos(250)
 * next equation...
 * may=Fksin(0)+Fnsin(90)+mgcsin(250)
 * m(0)=Fn+Wsin(250)=Fy
 * use to find Fnet: Fnet=sqrt(Fx^2+Fy^2)

c). What value of F (force) is required to move the sled up the plane at constant velocity?
 * Fnet=ma
 * acceleration is zero because the object moves at constant velocity
 * sum of the forces is zero
 * Fnet= 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? (b) with a speed 8 m/s while decelerating at a rate of 2.4 m/s² Visited the page, and will do again -- work is incomplete !
 * Draw a diagram.
 * when the object is not moving Fn=mg
 * solve for mass
 * solve for Fn using the sum of all the forces.
 * Fy= ma(y)
 * acceleration is 0
 * Fy= m(0)
 * mgsin(270) + Fn sin(90) =0
 * -mg+Fn=0
 * mg=Fn
 * Use the mass you solved for in A and the same equation
 * acceleration is -2.4
 * mgsin(270) + Fn sin(90) = m(-2.4)
 * -mg + Fn = m(-2.4)
 * Fn= m(-2.4) + mg

__**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.** initial speed?
 * becuase the bird is flying along a horizontal path, the y component of the initial velocity is 0.
 * use y=Voyt+1/2at^2 solving for t
 * once you have t use the equation x=Voxt + 1/2at^2
 * 1/2at^2 cancels out because ax= 0
 * Solve for Vox
 * 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?
 * knowing that Voy= o use y= Voyt + 1/2at^2 and solve for t
 * once you have t solve x using x=Voxt

__**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?
 * find the components of the initial velocity by using Vox=Vo*Cosθ and Voy=Vo*Sin
 * from there you solve for t using the equation x= Voxt+1/2at^2
 * the acceleration of the x componenent is zero so 1/2at^2 cancels out
 * rearrange the equation to get t=x/Vox
 * y is the height of the buliding
 * the acceleration of y is acceleration due to gravity
 * use Δy=Voyt+1/2at^2 and solve for y
 * Professor's Note : Very GOOD - visit my page for more hints ! **
 * Professor's Note : Very GOOD - visit my page for more hints ! **

__**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** ii. Find the magnitudes of **A** and **B** iii. What is the angle between **A** and **B**
 * to find **A** add (**A**+**B**)+(**A**-**B**). Your solution will be 2**A**. DIvide by two and you will have vector **A**.
 * Using vector, **A** substitute and solve for vector **B**.
 * Using vector **A** and Vector **B** from the previous part solve for their individual magnitudes using the pythagorean theorem.
 * To get the angle between **A** and **B** use inverse tangent.
 * Take the inverso tangent of (j^/ i^).

Professor's Note : Job well done ! check the answer for (iii) in my page


 * **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?**
 * Vo= 0**
 * V = 305m/min**
 * a= 1.22 m/s^2**
 * x=?**
 * **final velocity needs to be converted from m/min to m/s**
 * **305m/60s**
 * **the "new" final velocity is 5.08333m/s**
 * **Use V^2=Vo^2 + 2a X and solve for X**
 * **V^2-Vo^2/2a = X**
 * **X = 10.59 m**
 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest?**
 * Vo= 0**
 * V = 305m/min**
 * a= 1.22 m/s^2**
 * x=190**
 * t=?**
 * have to find how long it takes to reach max speed and decelerate back to rest (the times will be the same)
 * Use V=Vo + at
 * V-Vo/a =t
 * t1=t2=4.16 s
 * subtract the distance it takes to accelerate and decelerate from the total distance.
 * 190 -(2*10.59)
 * x= 168.82
 * solve for the remaining distance using V=d/t
 * d/V=t
 * t= 33.21
 * time starting and ending at rest= 41.53s


 * --** **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.**
 * x=50**
 * a=9.81**
 * Vo=0**
 * t=?**
 * i). How long does it take for the first stone to reach the water surface?**
 * use the equation, X=Vot = 1/2at^2 and solve for t
 * t= 3.19 s
 * ii). What is the initial speed of the second stone?**
 * **because we know that the second stone was thrown a second later but reached the water at the same time, t for the second stone = 2.19s**
 * **want to find initial velocity but we don't know the final velocity**
 * **use X=Vot +1/2at^2 and solve for Vo**
 * **Vo=12.10m/s**

Professor's Note : Good effort - Like the explanations A LOT !! **Do not include numerical values in the final answer**-- check my page and follow the style in the future