<p>Would anyone like to help me on this problem - pleassse!</p>
<p>A 225 kg crate is pushed horizontally with a force of 710 N . If the coefficient of friction is .20, calculate the acceleration of the crate. </p>
<p>?????????</p>
<p>OK, this is a Fnet equation.
So to start our formulas are:
Fnetx = Fapp- Ff = MA
Fnety = Fg - Fn = MA
Uk(coefficient of friction) = Ff/Fn = .2</p>
<p>So, first we must find Fn.</p>
<p>Fnety = Fg - Fn = 0, MA equals zero here because we arn't moving up or down on the vertical plane.
Fg = Fn
(9.81)(225) = Fg
2207.25 = Fg = Fn</p>
<p>.20 = Ff/Fn
.20(2207.25) = Ff
441.45 = Ff</p>
<p>Fnetx = Fapp - Ff = MA
710 - 441.45 = (225)A
268.55 = (225)A</p>
<p>A = 1.2 m/s^2</p>
<p>thanks so much!!!!!!! Would u like to help me with another one?</p>
<p>If I have time, but I have some work to do atm.</p>
<p>thanks barracuda, i really appreciate it (if u ever need help with anything, u can surely pm me) ... i just cant get this force stuff =( ... anyway..</p>
<ol>
<li>A sled of 50kg is pulled along a flat, snow-covered surface. The static friction coefficient if .30, and the kinetic friction coefficient is .10</li>
</ol>
<p>A. What does the sled weigh?</p>
<p>B. What force will be needed to start the sled moving?</p>
<p>C. What force is needed to keep the sled moving at a constant velocity?</p>
<p>D. Once moving, what total force must be applied to the sled to accelerate it to 3.0m/s^2. </p>
<p>For you , this is prob very simple - for me .. its rocket science.... thanks again</p>
<p>As my teacher would say...DRAW A FORCE DIAGRAM!</p>
<p>grr, he's annoying.</p>
<p>Ummmm, as far as I can tell, it says that the sled weight 50 kg... so A's already answered.</p>
<p>I would do the rest, but we did this stuff a while ago and I don't remember the exact formulas. The normal force is 9.8(50); that's a good place to start...</p>
<p>A) 50(9.81) = 490.5 (This will also be our Fn, same reasonings as before)</p>
<p>B) Fnetx = Fapp-Ff = MA
Ff/Fn = .3
.3(490.5) = 147.15 = Ff
Fapp - Ff = 0 (constant acceleration makes MA 0, this always applies when one says "Force to get object started" or "constant acceleration")
Fapp = Ff = 147.15</p>
<p>C) Ff/Fn = .1
.1(490.5) = Ff = 49.05
Fapp-Ff = 0
Fapp = Ff = 49.05</p>
<p>D) Fapp - Ff = MA
Fapp - 49.05 = (50)(3)
Fapp = 199.05</p>
<p>To Lavender: Weight is not mass. The answer is not given, you must multiply the mass by the constant of gravity to acquire weight.</p>
<p>I can try to help you with this one.</p>
<p>A. Weight=Mass*Accel. due to gravity=(50kg)(10m/s^2)=500N</p>
<p>B. The force must be greater than the force of the static friction. Thus, the force must be greater than Weight*Coef. if static friction=(500 N)(.3)=150 N.</p>
<p>C. This is the Same as (B) but with the coefficent of kinetic friction => F=(500N)(.1)=50 N</p>
<p>D. This is pretty much the same as the first question that you asked, only it is asking for a different variable. First we must find the net force on the object. Call the force that we will be applying F. F<em>net=F-50 (F minus the force of kinectic friction, which we found in part C.) F</em>net=m*a. We know m=50kg, and we want a=3, so we solve: F-50=50(3) => F=200N.</p>
<p>I hope that helped; it is hard to explain these with words.</p>
<p>Edit: gah, I am too lazy to delete this....</p>
<p>barracuda...right, forgot. This would be why physics is my lowest grade: I don't pay attention to little things. Or my teacher. Lol. But at least I pulled off a B+, and a 38/40 on my last test...</p>
<p>you guys are helping me soo much its not even funny! I know i keep asking u questions, but my physics teacher is a pyscho who grades these questions !!! Some are 30 points each =( .. and since u guys are smart - i can trust you ..... so here goes another one ... my teacher labeled this one "easy".. its in a foreign language to me =/</p>
<ol>
<li>The instruments attached to a weather baloon have a mass of 5.0 kg. The baloon is released and exerts an upward force of 95N on the instruments. </li>
</ol>
<p>A. What is the acceleration of the balloon and instruments?</p>
<p>B. After the balloon has accelerated for 10.0s, the instruments are released. What is the velocity of the instruments at the moments of their release?</p>
<p>C. What net forces act on the instruments after the release?</p>
<p>D. When does the direction of their velocity first become downward?</p>
<p>Here's some advice, get the princeton review SAT physics book. It got me a 760 when I went into the book only knowing a quarter of the material. Ha. (I studied it for only a week)</p>
<p>thanks for the advice jack ... i definetly need something to help .....</p>
<p>Vinny, your textbook should probably help too ;)</p>
<p>my textbook is horrible - thts why i usally come here because u guys explain it better than the freakin author... i hate my physics book</p>
<p>The Princeton Review Book was much better for what I needed than the textbook.</p>