Fahed+Almosa


 * __ Week 09: __**

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

__________________________________________________________________________________________________________________ __ 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? __
 * 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 coefficients of static and kinetic friction between the inclined plane and the sled are 0.25 and 0.15 respectively. __

__** A) ** ** We know that fs = fn * Ms, so we can find fn and by using this equation fn = w*cos(θ), W must be in Newton. For Ms, it’s already given in **__ __** the question. After we find fn, we should use this equation to find the least magnitude **__ __** ∑fx = 0 = F + fs – m*g*sin(θ) **** à **** ∑fx = 0 = F + W * cos(θ) – m*g*sin(θ) **__ __** The last step is solving to find for F **__

__** B) ** ** To find the min force, we can use this equation: **__ __** ∑fx = 0 = F - fs – m*g*sin(θ) **** à **** ∑fx = 0 = F – W * cos(θ) – m*g*sin(θ) **__ __** Then solving for F which is the minimum force. **__

__** C) ** ** First, we know that the constant velocity = 0. Then we should use the coefficient of kinetic friction instead of the coefficient of static **__ __** friction, so the equation would be: **__ __** ∑fx = 0 = F - fk – m*g*sin(θ) **** à **** ∑fx = 0 = F – W * cos(θ) – m*g*sin(θ) **__

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

__ We have to remember that the tension = m ( g + f ) if the acceleration in positive, and __

__ tension = m ( g – f ) when the acceleration in negative ,so __

__ A) F = 0 that means we should use the first rule which is tension = m( g + f ), __

__ so t = m(g+0) = m*g = 65N __

__ B) F= -2.4. For this one we should use the second rule because it’s accelerating downwards, __

__ So, t = m ( g - f ) = (65/9.81)*(9.81-2.4)= 42.2997N __

__** Professor's Note : Good Job! learn the method and understand it and avoid numbers and work with symbols **__ __** **__

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

__** a) ** ** Ay= 9.8, t=?, Δy=5.4 **__ __** The first step is finding the time by using this equation ((Δy)= Voy * t + ½ * a * t^2 ) **__ __** Then Δy = ½ a t^2. **__ __** (We canceled out the first part because Voy= 0). **__ __** After that, we find t, t=sqrt(Δy*2/a) **__ __** After we find the time, we can find the initial velocity by using this equation ( Δx=Vox * t + ½ * a * t^2) **__ __** The second part will be canceled out because the acceleration=0 **__ __** So it will be Vo=Δx/t **__

__** b) ** ** From part a, we can use the initial velocity but we have to find the new time. **__ __** Δy= Voy* t + ½ * a * t^2 ( Voy*t will be canceled out because Voy=0) **__ __** After we find the new time, we can solve for the horizontal displacement which is **__ __** Δx= Vox * t +1/2 * a * t^2 (1/2 * a * t^2 = 0 because the acceleration = 0) **__
 * 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?


 * Θ= 54 ° ,Vi=10.5m/s, A= h=0 ** (explain please **  ) , v=9.81m/s^2 , Δx=15.3m **

**Professor's Note: good, pay attention to the comment in red, also visit my page for more hints for tomorrow Quiz**
 * First, we have to draw a projectile graph to state the data given. **
 * From the formulas that we learned in class, we can use this formula to solve the problem: **
 * Δx = Vx * cos(Θ)* t - ½ * g * t² (don't use "g" in ** Δx equation) **, we should cancle out the second part becuase a=0, so it would be Δx = Vx * cos(Θ) **
 * After that, we have to rearrange the equation to get the time t = Δx / Vox **
 * In the graph that we drew, the height of the building is Δy, so the equation of Δy is **
 * Δy = Vy * sin(Θ) * t – ½ * g* t^2 **

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**
 * __ Week three (06th of February, 2012): __**
 * We have to add (A+B)+(A-B) to cancel out the B. **
 * A+B = 6.0 iˆ + 1.0 jˆ. **
 * A-B = - 4.0 iˆ + 7.0 jˆ. **
 * 2A = 2.0 i^ + 8.0 j^ **
 * Then we divide both sides with 2 to find A. **
 * A=1.0i^ + 4.0j^ **
 * Then to find B, we plug in A in any of these two equations and solve it for B. **
 * B= 5 iˆ - 3 jˆ **

ii. Find the magnitudes of **A** and **B**
 * -When we make triangle, A and B are given in the sides **
 * -We should use the Pythagorean theorem to find it: C=sqrt(A^2+B^2) **
 * |A| = sqrt( (1)^2+(4)^2) = sqrt(17) **
 * |B| = sqrt((5)^2+(-3)^2) = sqrt(34) **

iii. What is the angle between **A** and **B**
 * we can find the angle between A and B by using the dot product. **
 * The angle = tan^-1(B/A) **

**Professor's Note : Good job on part i and ii. For part iii - check my page**