Recent content by The Merf

ooh, and this one look very promissing: x(t)=\frac{v_0}{\omega}\sin(\omega t)+x_0\cos(\omega t)\;. I know that v0 is 0 so that wipes out the first half, I think x0 is 12.5, but where does the 4.9 come in and what is the w? hmm, the image doesn't seem to work so...

I also found the fallowing equations dealing with harmonic oscillation: x(t)=A\sin(\omega t)+B\cos(\omega t) \omega^2 =\frac{k}{m} i'm not sure what the w is

A mass on a spring, trying to find dispacement after at certain time. (Please Help!) Homework Statement A 0.27 kg mass is suspended on a spring that stretches a distance of 4.9 cm. The mass is then pulled down an additional distance of 12.5 cm and released. What's the displacement from the...

ok so N=mgcos(theta) N=79.1x9.8xCos(38) N=610.85 so friction=uN friction=-0.09x610.85 friction=-54.98 so the net forces N+friction=610.85-54.98 or 422.27 so that is the net force now K=mv^2 w=dK=(K-K_0)/2 so 1/2K (because he starts at rest so v_0=0 so K_0=0) or (mv^2)/2 now...

ok thanks

oops... my bad, i did use sin i just accidentally wrote cos so is it supposed to be cos?

ok so N=mgcos(theta) N=79.1x9.8xCos(38) N=477.25 (i think it is a little different because of rounding earlier) so friction=uN friction=-0.09x477.25 friction=-42.95 so the net forces N+friction=477.25-42.95 or 434.30 so that is the net force now K=mv^2 w=dK=(K-K_0)/2 so 1/2K...

I can complete this problem without friction, or even with friction, but in one dimension, but what confuses me is that usually gravity and friction work together to slow down an object, now gravity is working to speed it up and friction is trying to slow it down, so I'm not sure what equation...

ok since i got 48.69 to be the mass in the y direction, i take that times -9.8 as the acceleration due to gravity and get -477.16 as the force, i think i Newtons, but where does friction come in?

ok guys, i got one that wants me to figure out how much work is done: A skier of a mass 79.1 kg, starting from rest, slides down a slope at an angle of 38 degrees with the horizontal. The coefficient of kinetic friction, u, is 0.09. What is the net work (the net gain in kinetic energy) done on...