<p>Hey guys, im going to take the F=MA exam soon and am trying to study for it. My A.P. Physics B. class has covered most of mechanics (except for Torque and rotational stuff like rotational acceleration). I assumed it couldn't be that hard but going through it, i realized that it is indeed very difficult. The questions that i do get, take far too long for me to do, and the ones i think i get, i just don't get, and the ones i don't get, i just don't get it. I was going through the past exams and there was simple stuff that felt so complicated. </p>
<p>There was a time versus position graph(Looked like a cos wave stating at 0). They told me to find the total energy. I was stumped. </p>
<p>Anybody have any ideas how i can prepare with 5 days left. How can i learn rotational dynamics and torque quickly. How can i speed up solving.
Also if anybody can help me with the energy question, i would appreciate it.</p>
<p>Well, the equation seems to be x = 5sin(t<em>pi/4), so to find the velocity take the derivative so then v = 1.25pi</em>cos(t<em>pi/4), so in order to find the maximum velocity, then we need to set acceleration t=0, which is a = -1.25/4</em>pi^2<em>sin(t</em>pi/4). So in other words, the maximum velocity is 1.25pi<em>cos(0</em>pi/4), which equals to 1.25. So using the law of conservation of energy, then the maximum energy equals (m(1.25pi)^2)/2, which is about 7.7 times the mass. But I can’t seem to find where the mass is given.</p>
<p>exactly. i have no clue what to do. And even still i haven’t taken calculus so i wouldn’t be able to do this anyways. I’m just about to give up on this. Thank you anyways,</p>
<p>It’s -5 J. I have no clue why it’s -5 exactly, but maybe it has to be negative. If you focus on the beginning of the graph, the average velocity is 0. And since there is a change in position ok 8<t<10 the energy has to be negative…?</p>
<p>Fr 0<t<8 the average velocity is 0. Displacement over time gives you the average velocity. The displacement doesn’t change (well it does…but the difference is 0) in that interval therefor the average velocity is 0. </p>
<p>In the last sentence, I was just trying to think of anything that would justify the answer being negative. If energy is leaving the system, the energy is negative.</p>
<p>You have to use the graph from #18 where it says “The following graph of potential energy is used for questions 18 through 20.” </p>
<p>I think you’ll get it now but I will try to explain. I took this stuff last year so my physics is a little rusty:</p>
<p>Total energy of a particle = Kinetic Energy + Potential Energy
or Enet = KE + PE
Enet = (1/2mv^2) + PE</p>
<p>Using the graph of x vs. t in #20 we can find v using the slope of the graph (x/t). Where the slope of the graph = 0, the velocity = 0, and therefore the kinetic energy = 0.
This occurs at x= +/-5</p>
<p>So, at x = +/-5:
Enet = PE
Now, since there is no kinetic energy, you can look at the graph above #18 and see that the PE at both x = 5 and x = -5 is -5 Joules, choice A.</p>
<p>hmm ok but how did you get slope/velocity as 0 at ± 5. Delta x is 5 or -5 and delta T is 2, 10, or 6. The only way you cna get slope to be 0 is if delta x or delta t is 0 which only occurs at t=4 or 8.</p>
<p>At t = 2, 6, and 9 delta x = 0 because x is staying constant. At t = 4 and t = 8, x (position) is 0, but delta x (displacement) is changing. </p>
<p>It’s hard for me to explain but you’re reading the graph wrong. Just think of it this way, the slope is zero when the curve makes a horizontal line.</p>