Recent content by whynot314

Ahhh ok thank you, I thought I could only assume constant acceleration if the only force acting was gravity?

ok, so I take it that means IMMEDIATELY after impact thus. v=\frac{Ma\sqrt{2gr}}{(Ma+Mc)}

by included i mean that, I use the equation that is written with the friction force integrated with respect to time.

say a bullet with mass mb and initial speed v0 strikes and becomes embedded in a block of mass mc, which is initially at rest. The coefficient of kinetic friction between the bock and the surface is uk. ( a situation I set up) my question is, because its an impact problem. Can I ignore the...

yea nothing was given about a radius of curvature or anything. But to answer your question it would be v*dθ. (there is nothing about dθ in this problem either) I first thought that I could try to find a Normal velocity to the path, but quickly realized that there is no such thing lol. I...

A particle moves along the curved path sown. if the particle has a speed of 40ft/sec at A at time ta and a speed of 44ft/sec at B at time tb, determine the average values of he acceleration of hte particle between A and B, bath normal and tangent to the path. I found the average tangent...

Yea I saw those equations else where, but I wanted to use the method the book was giving me.

got it guys finally!, Yes you were right haruspex I was having HUGE difficulty as to what was being meant by "X". Thanks for the help haruspex. http://i1341.photobucket.com/albums/o745/nebula-314/20131226_161422_zps90413be0.jpg

This time I used the position of 8 in as x=0, which allows me to generally set up an integration for velocity where initial position is zero. does this look correct?

Ok my new strat, a= \frac{-2400x}{7}+4800 (general integration setup \int^{V}_{0}vdv=\int^{X}_{0} (\frac{-2400x}{7}+4800)dx gives me V= \sqrt{\frac{-2400x^2}{7}+9600x } then, velocity as a function of position. \int^{t}_{0}dt=\int^{3}_{0}\frac{dx}{ \sqrt{\frac{-2400x^2}{7}+9600x }} put...

yea idk I took x to be in the positive direction to the right, thus giving me a negative X. I want to use the equation of acceleration as a function of position a=f(s). and. \int^{V}_{V0}vdv=\int^{S}_{s0}f(s)ds to derive me velocity as function of position then, use that to further drive...

the 14 in. spring is compressed to an 8 in length, where it is released from rest and accelerates black A. the acceleration has an initial value of 400ft/sec^2 and then decreases linearly with the x-movement of the black, reaching zero when the spring regains its original 14 in. length...

But your solution did help a lot thanks.

Oh i was trying to derive that equation, and perhaps save some steps.

v2 being initial velocity.