Okay, so, correct me if I'm wrong:
The energy equation that should be used is KEi+PEi=PEf+KEf; However, there is no KE at the maximum amplitude (because it comes to rest), so the equation should thus be KEi+PEi=PEf.
Well, KEi+PEi = .5*m*v^2 + .5*k*x^2 right?
I assume I use the mass of both...
I don't understand this completely. Why isn't the energy of the clay conserved during the collision? Ei=KEi+PEi, right? And doesn't Ei=Ef?
v= (mvi+Mvi)/(m+M)
This equation gives the final velocity of 2 objects, of mass m and M with intial velocities vi1 and vi2, the collide and stick. It...
I have been working on this problem for an hour and can't seem to figure it out. I know there must be something fundamental that I must be doing wrong. Can someone give me a pointer?
A platform of mass 0.8 kg is supported on four springs. A chunk of modeling clay of mass 0.6 kg is held...
Hi, I have worked through this problem for so long that I cannot see how I could be getting it wrong. I feel like I am doing it correctly, but I don't get the right answer.
QUESTION
A disk of mass M1 = 350 g and radius R1 = 10 cm rotates about its symmetry axis at finitial = 154 rpm. A...
Is there something I am completely missing here?
dx=vo*t+(a*t^2)/2
20.2 = 19.9t + 9.81/2*t^2
solving for t = 0.84 or -4.9
2*t = 1.68s. Wrong.
Please, does anyone have any suggestions?
Thank you so much.
Hi:
I am working through this problem regarding a bouncing ball. I have gotten most of the way through, but can't figure out the time interval between bounces. Can someone give me some guidance? What am I not seeing?
A 32 g steel ball bounces elastically on a steel plate, always...
First off, sorry I forgot - yes, the height of E is the same as D, 0.
So, I did as you said:
KE = mgy = µmgx
where KE is the kinetic energy at D, y is the height at A (50m), µ is the coefficient of kinetic friction (0.4) and x is the distance traveled to stop.
Solving for X...
I am now even more confused than I was. What about friction? If you are saying that KE at B is the same as at D, then it'll be
KE=mv^2/2=(220)(31.32^2)/2 = 107903.66 J.
Okay, that makes sense, I guess. Then we have to consider PE, which somehow you all found to be 105600 J, but I don't know...
I feel like a fool because I can't get this, but I am simply in need of some guidance re: equation usage.
A bobsled run leads down a hill as sketched in the figure above. Between points A and D, friction is negligible. Between points D and E at the end of the run, the coefficient of kinetic...
I will look into the tractor problem. Would the gravity force incorporate weight (mg)?
As for the other problem (these are from the book - I am "practicing", albeit unsuccessfully): "A 65-kg skier speeds down a trail. The surface is smooth and inclined at an angle of 22 deg with the...
Right, I understand what you are saying. However, this is a part in the book BEFORE any mention of weight or normal forces. So they cannot be incorporated into solving the problem. So I feel like the only force at play here (given that fact) is the force of the farmer. Right?
A similar...
I am finding myself stuck on what is likely considered a very basic problem. I know the answer (it's in the book) but after 20 minutes of searching and trying to find a method, I've turned to you. Could someone please describe the methodology for this?
A farm tractor tows a 4300kg trailer...