<p>Acceleration can’t be g/2. It changes and approaches g until the entire rope is over the edge. I assumed there was no tension force though. I think you were supposed to factor it in.</p>
<p>F = Mnet * a = (M/L)y * g</p>
<p>a(y) = (yg)/L</p>
<p>v(y) stumped me until after I got out. You have to use conservation of energy and use a reference frame where the height is equal to the entire length. That’s because once the entire rope has fallen off the acceleration will be a constant g.</p>
<p>Yea, I know I put that one of the velocities was greater, I forgot which one though. I prob got 25/45 total points on the free response for Mech.
For E/M I totally failed. #1 I got really long equations after find the derivative…
and my graph was an upwards parabola to the left of the line, and it increased at a decreasing rate to the right…</p>
<h1>2 I just used formulas and plugged in the numbers, hopefully I did it right.</h1>
<p>B was like 400 T right?
The bulb 1 is just as bright as before in #3 right?</p>
<p>When you solve the 2nd order DE, how do you get it to the form Asin(wt + T) or anything like that. In my DE class, we learned producing two linearly independent solutions from the characteristic equation.</p>
<p>It was the same brightness for D, when you added bulb 3. It should be less for E when you split the current square.</p>
<p>For #3 mech. </p>
<p>a. t = (m/2)a
(m/2)g - t = (m/2)a –> a = g/2
so v = sqrt(gd)</p>
<p>but how do you get c/d?</p>
<p>For the E & M</p>
<p>1a) i think i had radially inward for r < R and radially outward for r > R since the first E field was a negative function of r and the second was positive</p>
<p>1d) the graph was linearly increasing (starting at the origin) until R, where it started decreasing exponentially with a horizontal asymptote at the x-axis? i think the equations i derived in the earlier parts supported this…</p>
<p>2b) i drew lines going horizontally from a to b</p>
<p>2e) i drew lines going vertically up</p>
<p>3d) i said the bulb got brighter because having the resistors in parallel decreased the total resistance in the circuit (resulting in an increase in current/brightness)</p>
<p>3e) wasn’t really sure but i said they stayed the same since the size of the curcuit and the resistance of the bulbs was unaffected.</p>
<p>Axeback, do you even read?</p>
<p>since u = 0, THERE IS NO Ff…
I’m not saying I did it right, but that’s the only way I see logical…</p>
<p>@ sat2500</p>
<p>c) was Fg(found in b) x y, that gives you work.
d) I did Work = change in K.energy –> Work (partc) = 1/2mv^2 (since Ki = 0)
from there I got v= sqrt(2gl)</p>
<p>That is what I did, but doesn’t that only hold for constant force? I’m told you need to integrate the part b result</p>
<p>Pretty sure the velocities are equal in number 3.</p>
<p>One had speed as a function of R, while the other had it as a function of sqrt(R). Therefore, the first is faster.</p>
<p>For #3 I had a. v = sqrt(gd) c. v = sqrt(2gL) and for part e. I had the velocity of the rope is greater by a factor of sqrt(2).</p>
<p>Also, does anyone remember their value for momentum and acceleration for mechanics part 1?</p>
<p>A. I beg to differ. I find that the velocity of the hanging block was greater. No force of friction means no tension…or at least so I thought. It ended up being an (M/2)gd=1/2(m/2)v^2. Once you solve that you get v=sqrt(2gd).</p>
<p>B. For part B doing a sum of the forces results in (y/L)Mg. (Which is the function based on y for the FORCE of gravity on the object) </p>
<p>C. From there you can integrate that force function with respect to the y and that gives you the work. I came up with a ((y^2)/2L)Mg.</p>
<p>D. From this you can say that work is the change in kinetic energy where the kinetic energy is 0 initially. Thus, the final speed is sqrt((gy^2)/L).</p>
<p>Sharvas: For momentum, find the kinetic energy @ 0.60m then find the v of the kinetic energy. Take that v and multiply it by your mass. Acceleration…well, I took the negative derivative of potential energy with respect to position which is equal to force. I then divided by the mass yielding the acceleration. p=5.78 kg*m/s ; a=-1.6 m/s^2</p>
<p>E. From this you take the limit as y approaches L. you get a sqrt((gL^2)/L) which causes the denominator L to disappear resulting in sqrt(gl). Clearly the block has a greater acceleration due to the fact that it sqrt(2gL) is greater than sqrt(gL).</p>
<p>For E&M #1, can anyone confirm that r<r is=“” radially=“” inward=“” and=“” when=“” r=“”>R is radially outward?</r></p>
<p>I think I put inward, since the voltage was smaller in the center than on the surface.</p>
<p>The acceleration of the two block system can’t be equal to g. Only half of it has a downwards force on it. The block on the table has a counteracting normal force. </p>
<p>So you have 1 net force acting on a system of mass m/2, but the total mass of the system is m. Thus the net acceleration is g/2.</p>
<p>The difference between my work and your work is that you made the acceleration g. When I made it g/2.</p>
<p>The acceleration acting on the block by the table is acting on it in the y direction. The normal force doesn’t stop it from being accelerated, sulthernao. Do a sum of the forces and you’ll see. >.></p>
<p>Wait, why are you doing a sum of the forces? I just did the conservation of mechanical energy…</p>
<p>But doesn’t the gravity only “pull down” on the mass that is not on the table. The gravity on the other object is balanced by the normal force.</p>
<p>Fnet = ma = m/2*g -> a = g/2. </p>
<p>Maybe I’m wrong, but how can the acceleration equal to g in part a and then not equal to g in the rope situation. </p>
<p>You’re saying the the acceleration is g no mater what. Thus why doesn’t the rope constantly accelerate at g no matter how much of the rope is over the table?</p>
<p>There has to be tension in the rope because the block on the table is accelerating, and no other force could possibly cause that other than the tension. I really hope this argument is valid.</p>
<p>so block 1 has mg down, normal force up, and tension to the right. Tension exists despite the table being frictionless</p>
<p>I still don’t get why you’re doing the sum of the forces. How do you get velocity from that? And yes, the acceleration is constant. The acceleration is also g. It is balanced by the normal force which is why they asked you to create a function for the force of gravity on the rope. Gravity only accelerates the hanging part of the rope. The part that isn’t hanging is balanced, but the acceleration downward causes the balanced part to accelerate to the right. As far as the blocks go you do conservation of mechanical energy for the blocks.</p>
<p>sat2500: Well, I still don’t understand why you’re talking about forces and such. Just do a conservation of the mechanical energy about the system of the hanging block.</p>