Recent content by littlehonda

ok here's what I did integrated F with respect to r and got W W=-k/r + c1 using W = -\DeltaU I found U to be k/r I then used the expression \textbf{E} = \frac{k}{r1} +\frac{1}{2}mv1^{2} = \frac{k}{r2} +\frac{1}{2}mv2^{2} and since v1= 0 and r1=0 \textbf{E} = \frac{k}{r1} =...

K is given as a constant, analogous to Coulombs or the gravitational constant. The force could be expressed as F = -k/r^2 = m\ddot{x} and we can see that acceleration is as always inversely proportional to the mass of the particle. I believe you need to use U to solve this problem but...

Homework Statement a particle of mass m is attracted to the origin by a force F=-k/r^2 find the time t for the mass to reach the origin Homework Equations \DeltaU= U-U0 = \int\textbf{F}(r) dr The Attempt at a Solution I found \DeltaU by \DeltaU= U-U0 = \intF(r) dr and by...

hey thanks a lot both of you guys! hate to keep pestering but if I want to find x(t) I just integrate dx/dt and find a new constant C3 using the initial condition x(0) t=0?

v= (ln(\alpha(-At+C3)))/\alpha Solving for C3 C3 = (e^v0\alpha)/\alpha at V(0) t=0 C3 = 73.89 v= (ln(\alpha(-At+73.89)))/\alpha

I was careless putting the force equation up, the right side of the equation also contains a m to cancel the one on the left.

v_0=v(t=0) (e^20\alpha)/\alpha = C3 C3 = 73.8 v(t) = (ln(-At\alpha)+ ln(73.8\alpha))/alpha sorry if I am a little slow with this, am I making an error here?

ok so that's where I got to pretty much, the C1+C2=C3 makes sense. Here is all the information given A= 1/m/s^2 \alpha=.1 s/m v0= 20 m/s

thanks for responding gabba, I think that I tried the separable differential equations method but couldn't make sense of the answer what i did was e^(\alphav) dv = -Adt 1/\alpha(e^(\alphav)-e^(\alphav0)) = -At i'm not sure what this means though or where to go from here

Homework Statement Given the one-dimensional retarding force F=-Ae^(-\alphav) find an expression for v(t). Homework Equations F = m(dV/dt) A and \alpha are constants, v is instantaneous speed. The Attempt at a Solution Im not sure how to frame the idea of integrating a...