RyanAmedeo

Week 9 Solutions __ a) __ Work= Force·distance*cos(theta)

The force needs to be found before moving on.

Force=mass*acceleration

Tcos(0) + mass*gravity*cos(180)=mass*acceleration

T= mass*gravity+mass*acceleration

Work=(mass*gravity+mass(gravity/4)·distance

b)

Work=Force·distance*cos(theta) Same thing as the first one, solve for f using F=ma before plugging into W=f*d*cos(theta)

Force=mass*gravity

Work=mass*gravity*distance*cos(0)

Work=mass*gravity*distance

c) Kinetic Energy =(1/2)* mass*velocity^2 Kinetic Energy=(1/2 )mass(sqrt(2(gravity/4)distance)

d)

velocity^2=initial velocity^2 +2*acceleration(delta x)

velocity= Sqrt(2(gravity/4)distance)

Week 8 Solutions a) There are three forces on the block- the normal force, the force of friction, and the force of gravity. In order to get the force of friction, I multiplied the the weight of the sled by the force of friction so its 85cos(20) Then you have to multiply this number by .25 because that was the amount of friction given. Then in order to find the for gravity and to do that you add 85sin(20), since it is in the y direction. Then once these solutions are found you add the two together to get your answer.

b)With your first answer you just have to plug any force larger than it.

c) This problem is the same as part a, besides if you plug in .15 instead of .25 so your equation looks like 85cos(20)*.15 + 85sin(20)

Week 7 Solutions a) Since the is no acceleration in the x direction the answer is 65 N.

b)First we must find the force Using the y direction, since there is no acceleration in the x direction I got: Fy=Ftsin90+mgsin270=ma Fy=Ft-mg=ma--> sin90= 1 and sin270=-1 Once moving around the equation to equal Ft, FT=ma+mg

Then to find the mass I did 65=m*9.8 m=6.63

then using the equation found above: The 2.4 is negative because it is decelerating. Ft=(-2,4*6.63)+(6.63*9.8) Ft=49 N

**Professor's Note** : Good job Ryan ! Keep up the good work. Although no comments are made on the last two week work, I have given you the extra credit points.

Week 5 Solutions

a)With the information given: DY=5.4 m a=-9.8 m/s^2 Vy=0 t=?

DX=8 m a=0 m/s^2 Vx=? t=?

Using the equation and the y numbers I solved for t.

DY=Vot+.5at^2 5.4=(0)t+.5(-9.8)t^2 t=1.04 seconds

Then once t is found using the same equation I solved for delta x.

8=Vo(1.04)+.5(0)(1.04)^2 Vo=7.69 m/s

b) To find DX you first need to use the equation DY=Vot+.5at^2, using 2.7 as your new dy and solve for t.

Once t is found use the same equation but from the x perspective and solve for DX giving you how far the fish traveled.

Week 4 Solutions -First using the angle given and DX, I found Vx and Vy using cosine and sine. V0=10.5 <54 delta(x)=15.3

cos54=Vx/10.5

10.5cos54=Vx Vx=6.17

10.5sin54=Vy Vy=8.49 -Then using the velocity equation I found the projectile height of the rock from the top building. V^2 = V0^2 + 2a(dy) (8.49)^2=0^2 +2(-9.8)dy -72.08=19.6(dy) dy=3.67

-Next using the Vy found at first the time was found using the velocity equation of y. V^2 = V0^2 + 2a(dy) 0=8.49-9.8t -8.49=-9.8t t=.87 seconds -Next once the other parameters of x were found I plugged them into the equation to find dx.. dx=V0t+ .5at^2 dx=(6.17)(.87)+.5(0)(.87)^2 dx=5.37 meters -Then I subtracted the dx given in the problem minus the dx I had just found to get a total of 9.93 meters, which was needed to subtract from dy in the end 15.3-5.37=9.93 meters

dx=Vot+.5at^2 dx=6.17+ .5at^2 9.93=6.17t t=1.61 seconds -Using the time from the equation before, I found the delta y total which would be subtracted from the delta x total. dy=Vot+.5at^2 dy=0+.5(-9.8)(1.61)^2 dy=12.7 12.7-3.67=9.03 meters

**Week 3 Solutions**

1)If **A** is added to **B** the result is 6.0i^ P1.0j^. If **B** is subtracted from **A** the result is -4.0i^ +7.0 j^.

a) Find **A** and **B**

-In order to find **A** and **B,** I added (**A+B)** and (**A-B)** in which the **B**'s will cancel out and then all that is left is **2A.** -After this I divided both sides by 2 in order to solve for A. -Then if you plug **A** into (**A+B**) and then solve for **B.**

b)Find the magnitudes of **A** and **B**.

-I made two right triangles, one using for **A** and one for **B**. -then I used to pythagorean theorem which is sqrt(x^2+y^2)=(hypotenuse) -Doing this with both triangles will give you the magnitude for both vectors.

c)What is the angle between **A** and **B**?

-In order to find the angle, I used the trig identity using tangent. -For the purpose of this problem I will use T as my theta. -So then the equation to find the angle is tan(T)=(j^)/(i^) -In order to solve for T I used the inverse tangent of (j^)/(i^). -My final equation looked like T=tan^(-1)(j^/i^)

Professor's Note : very Good !! - Liked your explanations a lot ! I am not sure if part three is clear enough -- check my page

1) **PART 1** DX=? V=305 m/m ---> 5.0833 m/s V0=0 m/s a=1.22 m/s^2 t=?

V^2=V0^2+2a(DX) (5.0833)^2=0^2 +2(1.22)DX 25.84=2.44DX DX=10.59 meters 10.59=start and 10.59=stop--> 21.18m accelerating 190-21.18=168.82m at constant speed V=V0+at 5.08=0+1.22t t=4.17
 * PART 2**

speed=(distance/time) 5.08=(168.82/t) t=33.21 seconds + 8.34 seconds = 41.55 seconds

2)**PART 1** DX=50 m a=-9.8 m/s^2 t=? V0=0 DX=V0t+.5at^2 50=0t+.5(-9.8)(2.19)^2 50=4.9t^2 10.2=t^2 t=3.19 seconds

50=V0(2.19)+.5(-9.8)(2.19)^2 50=V0(2.19)-23.5009 73.5009=V0(2.19) V0=33.56 m/s
 * PART 2**

Professor's Note : very Good effort - Like to see the steps and explanations !! no need to include numerical values or the final answer -- check my page