Philip+Gaudio

__**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**

i. Ax+Bx= 6i & Ax-Bx=-4i Therefore **Ax= 1i & Bx=5i** Ay+By= 1j & Ay-By=7j Therefore **Ay=4j & By=-3j**

ii. Magnitudes would be calculated by finding the hypotenuses of the right triangles. For **A**, it would be 4.12311 and **B** would be 5.83095.

iii. A. B= -7 abs(A) * abs(B) =24.04

Since A. B= abs(A)abs(B)cos(theta) then,

cos^-1(-7/(the product of the magnitudes)) =theta which equals 106.928 degrees.

Professor's Note : Job well done !!

__**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?

15.3= deltax Vix=10.5 theta=54 a=9.8 X component of Velocity= 10.5cos(54)=6.17 Y component of Velocity= 10.5cos(54)=8.49

The vector has to be broken down into proper components.

15.3=6.17t so t=2.48 seconds is the time this takes.

deltay=8.49t+(.5)(-9.8)(2.48)^2=-9.08

The answer will be positive 9.08 meters because the projectile is going below horizontal axis of the building, therefore it falls the equivalent of the height of the building.


 * Professor's Note : ** Good job -- don't include numbers ! visit my page for more hints

__**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 should set up a table with all known values and take note of what we are trying to find. a) vf2 = vi2 + 2a(deltay) t, This equation will give the proper vertical velocity. Now we need to find time using delta x = 1/2(vi + vf)t From this equation we can now use delta x = vi(t) + 1/2a(t)^2 to figure out the initial velocity of the fish. b) We would use the same three equations but this time we are solving for delta x and not vi.It would involve first finding Vf, then time, and finally deltax.

__ **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) Since the velocity is at a constant pace, no outside Force is acting upon the object. So, T=m(g+0) and so the person's weight 65N. (b) Since the velocity is unchanged, but there is a slight deceleration while the cab is moving upward, it is essentially moving up at a slower and slower rate. This means that the scale would read LOWER than 65N. The new outside force f=2.4 so T=m(g-f) would result in reading below 65N.

**Professor's Note :** Good Job, PJ ! Make sure that you use the absolute value for f now, as you have already taken care of the negative sign !

__** 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

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?

a). So, the first thing we should do is solve for F by setting its equation equal to zero. F+weight*cos(theta)*(static friction coeff)-weight*sin(theta)=0 Solve the equation for F.

b). The minimum Force would be the same equation but it would start with F- instead of F +.

c). This would be the same as the first equation using kinetic friction coefficient instead of the static one.

Week 9 (April 26th, 2012)

Draw a picture. A. Find the work done by the cord's force on the block. B. Find the work done by the gravitational force on the block. C. Find the Kinetic Friction. D. Find the speed of the block.
 * 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,

A. The net force would be mg/4 downwards, while weight would just be mg. T is equal to the force of the cord on the block. T=m*(g-g/3) W=F. d cos(theta), W=-T. d. T is negative because the force's direction is against the motion's.

B. W= M*g*d

C. W= delta K so W= number 2's answer minus number 1's answer.

D. K=(1/2)m*v^2, solve for v.

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.

A) So mgh = (1/2)mv^2 & v = sqrt(2*g*h). So the speed of the ball would be Vf1= (M1-M2)*sqrt (mgd)/(M1+M2).

B) Now we're just going to solve for Vf2=(M2(M1)/(M1 + M2))*sqrt(mgd).