Justin+Kenefick

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) mgh= 0.5(m1)v^2**
 * gh= .5 v^2**
 * solve for v1**


 * b) m1v1 = m2v2**
 * m1 and v1 are now known, as well as m2.**
 * solve for v2**

At first I thought you had the right thing but I am afraid its not the case ! please refer to my page-- professor

__Week 9__


 * 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·dcosƟ**
 * Need to find F**
 * F=ma**
 * Tcos(0) + mgcos(180)=ma**
 * T= mg+ma**
 * W=(mg+m(g/4)·d** answer in joules

B. Find the work done by the gravitational force on the block W=F·dcosƟ F=mg W=mgdcos(0) W=mgd

D. Find the speed of the block V2=Vo2 +2a(Δd) Vo is zero V= Sqrt(2(g/4)d)

C. Find the kinetic energy of the block KE =1/2 mv2 Use the v from part d KE=1/2 m(sqrt(2(g/4)d)


 * __ 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) -find the normal force of the sled to plug into the static friction equation Fn = mg(85) cos(20) -plug into static friction equation to find Fs Fs=0.25* Fn

-find the x component of the gravitational force 85 sin(20) -since the static friction and the gravitational are opposite directions, find the extra F needed to keep it in place by finding the different between them Fgx-Fs= 9.1

B) minimum force >9.1

C) same thing, except with the kinetic friction value of .15 instead of the static friction of .25


 * __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) -find the mass of the hanging object. Since the force on the spring balance is 65N, i knew that that is equal to 9.8M. therefore to find the mass i divided 65 by 9.8. -The i found the net force of the hanging object and set it equal to MA. The tension was in the positive direction and mass was in the opposite direction. To find the net force then, i used Ft-mg = ma -then i plugged in the mass, the acceleration was equal to zero (constant speed), and the gravitational constant, and solved for Ft

b) - i plugged in the same numbers, except with the acceleration at 2.4 instead of 0.

Professor's Note : Justin, you got the first part right, but the in the second part, a = -2.4 (as it a deceleration !)

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 Five February 24th__**

a) -find the time using the following equation and the initial velocity as 0. (delta-y)=Viy * t + (.5)(ay)(t)^2 -use the same equation but with the x components, using the time from last equation and solving for the initial velocity in the x direction. -Find the initial speed of the fish using trig and the initial velocity components (0 in the y, and the x you just solved for)

b) -solve for the new time in the same way as the first step of part A. -use s=d/t to solve for the new delta x of the fish, using the initial x velocity from part A, and the time just solved for.

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?
 * __Week four (13 th February, 2012)__**

1. separate the projectiles x and y components. -use cosine of the angle to find the Vi-x -use sin of the angle to find the Vi-y 2. Find the change in y from the top of the building to the peak of the projectiles path -i used the third equation (using gravitational constant, Vf-y=0, Vi-y) 3. find the time it takes for the projectile to reach its peak (using the change in y just found, as well as the Vi-y and the Vf-y - first equation 4. Find the change in x from the projectiles start to its peak height - second equation 5. subtract the change in x(beginning to peak) from the given total change in x to find the change in x from the peak to when the projectile hits the ground 6. Find the time it takes for the projectile to travel from peak to the ground using the second equation -Knew change in x, initial and final velocities (since x velocity remains unchanged throughout) and that ax=0. 7. find the change in y for the second part (from when the projectile was at peak to when it hit the ground. - i used the second equation as well as the time that was just found in 6, the gravitational constant, and the fact that Vi-y at peak was 0. 8. Subtract the change in y(from beginning to peak) from change in y(from peak to ground) to find the height of the building. - for the final height of the building i got 9.03 meters.


 * Professor's Note: Wow !! great job, also visit my page for more hints for tomorrow Quiz**

Week 3

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. - First I added **A** and **B.** the B's cancelled eachother out and left 2**A**. - I divided that result by 2 to solve for **A** - I plugged **A** back into the original equation to find **B**

ii. - set up two right triangles - the first uses the the two side lengths of **A** as the two legs. - the second uses the two side lengths of **B** as the legs -solve each for their hypotenuses using the pythagorean theorem to find the magnitudes of each vector

iii. - use any of the trig functions (all the side lengths are known) to solve the angles of the triangles using trigonometry. - add together the two angles that would properly find the angle between **A** and **B**
 * Professor's Note: Job very well done. For a different method to solve PART iii visit my page**


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


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


 * Vf= 305 m/min = 5.0832 m/s**
 * Vi= 0 m/s**
 * a=1.22m/s^**
 * x=?**


 * v^2 = vi^2 +2a(X)**
 * (5.0832)^2 = 0 + 2(1.22)x**
 * 25.84=2.44x**
 * x=10.59 m**


 * Vf=Vi+at**
 * 5.083 = 0 + 1.22 t**
 * 4.166 seconds = time to accelerate**
 * The time will be the same to accelerate and decelerate**


 * 10.59 m (to accelerate) + 10.59 m (to decelerate)**
 * 21.18m total**
 * 190m - 21.18m = 168.82m at constant speed of 5.0832m/s**


 * s=d/t**
 * 5.0832 m/s = ( 168.82 m / t )**
 * t= 41.54**


 * total time=**
 * 4.166 sec + 4.166 sec + 33.2127 sec = 41.54 seconds**

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? ii). What is the initial speed of the second stone?


 * first stone**


 * x= 50 m**
 * a= 9.81 m/s^2**
 * Vi=0 m/s**
 * t=?**


 * x=Vi * t + .5 at^2**
 * 50m = 0 + .5 * (9.81(m/s^2))*t^2**
 * 50 m = 4.9 t^2**
 * 10.2 = t^2**
 * t=3.19 sec (for first stone to hit water)**


 * second stone**
 * x=50 m**
 * a=9.81 m/s^2**
 * t = (3.19 - 1) seconds 2.19seconds**
 * Vi= ?**


 * x= Vi*t + .5 a t^2**
 * 50 = Vi (2.19) + .5(9.81)(2.19)^2**
 * 50= Vi 2.19 + 23.5**
 * 26.5 = Vi * 2.19**
 * Vi=12.1 m/s**


 * Professor's Note: good job, not necessary to include numbers !**