hassan-a

Hassan Alhafshan week 2
 * 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. **

* change the unit for speed ( velocity) from m/min to SI unit m/s 305 m/min * 1 min/60 sec = 305/60 = 5.08 m/s


 * Vo = 0 **
 * V= 5.08 **
 * a=1.22 **

by using the equation v^2=(v0)^2+2a(x-x0) (5.08 m/s)^2=(0 m/s)^2+2(1.22 m/s^2)(x-x0) (x-x0)= (5.08/m/s)^2/2(1.22m/s^2)= 10.58m
 * i). How far does the cab move while accelerating to the maximum speed starting from rest? **

v=v0+at. 5.08m/s=0+(1.22m/s^2)*t t= (5.08m/s) / (1.22 m/s^2) = 4.16 s **professor's** **note : second part, wrong use of starting and ending velocities and the sign for acceleration --> it is a deceleration**
 * ii). How long does it take to make the nonstop 190 m run, starting and ending at rest? **


 * 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-x0=v0*t+1/2*a*t^2 x-x0 = 50 Vo=0 t=? a= +9.8 50 m = 0*t + 1/2* (9.8m/s^2)* 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= 3.19 - 1=2.19 s Vo= ? x-xo=Vot+1/2at^2 50 m = Vo*( 2.19 s ) + 1/2 (9.8 m/s^2) (2.19^2 s) Vo= 12.1 m/s
 * How long does it take for the first stone to reach the water surface? **
 * What is the initial speed of the second stone? **

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.

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

we must first vertically allign the two equations. Then add the two equations to eliminate B and solve for A.

**A** + **B** = 6.0 iˆ + 1.0 jˆ ---(1) **A** - **B** = - 4.0 iˆ + 7.0 jˆ --(2)

ii. Find the magnitudes of **A** and **B** subtract the equations to eliminate A and solve for B

iii. What is the angle between **A** and **B** iii). we can find the angle by using the dot porduct between A and B.


 * Professor's Note : Be more specific on the steps. **

__** Week four (14th February, 2011) **__ 01. A 0.65 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?


 * we have to find the x and y components**

- ax = 0
-ay=9.8

=
- we have Vo= 10.5m/s we have to solve for t and the height =====

(x-x­o) = Voxt + 1/2at²
- after we get t solve for height (y-y­o) (y-yo­) = (Voy)t + 1/2(ay)t²

**Professor's Note: good work, visit my page for more hints for tomorrow Quiz**

**-**

__**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?

**first we have to find the time** **we have that :** . Voy = 0 m/s,

g = 9.81 m/s2 V² = Voy² + 2g(DeltaY), DeltaY = 5.4 m

we find time t.

V = V0y + gt to

after we find the time we find delta x after that we use the equation to find the initial velocity DX = V0xt the speed = V0x

b we Find Vy.

Vy² = Voy² + 2gDeltaY,



we find deltax and that gonna be the answer

__ **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 the speed is constant so the accelration =0 T = mg + ma ma=0 T=65 N b)  it is decelerating so a=-2.4  T-mg =ma  mg=65n  m=w / g  65=m/9.8  m= 6.6kg
 * Week 06 -- EXAM WEEK --- No wikis**

T = ma + mg =6.6*(-2.4) + 65 =49.16 N

Note; I sent you an email

Professor's Note : Good job !!

-
 * __ 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

-

Week 9 (April 26th, 2012)

Using F=M*a, consider a FBD of the block M*a=-M*g/4=T-M*g The tension in the cord is T T=M*g-M*g/4 T=3*M*g/4
 * 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,

The work done by T is -3*M*g*d/4

The work done by the Weight of the block is M*g*d the net is equal to the kinetic energy, or M*g*d/4

This is related to the speed of the block, v, as .5*M*v^2=M*g*d/4 v=sqrt(g*d/2)

j

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.

a PE b + KE b = PE a + KE a mgh+1/2mv1‫^‬٢=mgh+1/2mv2^2 gh=1/2v2^2 solve for v2 v2=sqrt(gh/2)

-‫‬ b PE b + KE b = PE a + KE a M1Vo1+M2Vo2 then we solve for V2f