Recent content by phazei

sigh. apparently the original position it was referring to was the equalibrium position, not the position it was initially pulled to. So cos has to be solved to equal 0. cos( (k/m)^(1/2) * t) = 0 t_1 = (pi/2)(m/k)^(1/2) I think the wording is poorly chosen.

the problem gives x(t)= A cos( (k/m)^(1/2) * t) A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. View Figure . Assume that the +x direction is to the right. The...

SHOOT! I asked it to just give me the answer so I could do the next part... There were a few other problems where it was saying it was at rest at a infinite distance... so I was still thinking that way... I was thinking K and U were 0 at an infinite distance... I totally forgot it had an...

I submitted the problem and it said it was wrong... :(

"I found m*b_max*v=m*R*v_f, I solved for v_f and put it into the E_surface equation and solved for b_max^2= 2*G*M_e*R/v^2" mrv... isn't this the conservation part of it? m*b_max*v = m*R*v_f the initial and final angular momentum...

In this problem, you will estimate the cross section for an earth-asteroid collision. In all that follows, assume that the Earth is fixed in space and that the radius of the asteroid is much less than the radius R of the earth. The mass of the Earth is M_e, and the mass of the asteroid is m. Use...

oh... mg=-kdeltaS 5(-9.81)=-490deltaS :biggrin: :biggrin: so they're different because it occilates when dropped and not when lowered, right?

the net force must be 0. So the force upward by the spring = mg But how to i relate that to figure out what the displacement is?

A vertical spring with k = 490 N/m is standing on the ground. You are holding a 5.0 kg block just above the spring, not quite touching it. A)How far does the spring compress if you let go of the block suddenly? B)How far does the spring compress if you slowly lower the block to the point...