Recent content by dontdisturbmycircles

I have found in the past that it greatly depends. There are times where I have gotten together with a few friends to study together, and we get absolutely nothing done -- we just chat. With other groups, we have gotten into a good rhythm and had a really solid review session. I think that group...

Start with the conservation of momentum, m1v1=m2v2 (the v's are vectors but since in this case its a linear problem it doesn't matter)

If he is walking up the steepest angle, he will just be on the verge of slipping, draw a free body diagram. The second part is more of the same feast.

Just cause something is difficult doesn't mean its impressive: case in point: Although I suppose some may find that impressive, lol!

np, don't be sorry! I just wanted you to understand that the only two forces acting on the person are the gravitational force which does work in the amount mgh=1300J, and the frictional force. Since we know \Delta E_k=W , we have 1/2mv_2^2-1/2mv_1^2=1300J - W_f where W_f is the work done by...

No, that's not right. The gravitational potential energy at the top is mgh=1300J The kinetic energy at the bottom is 140J The difference is 1300J-140J = 1160J Energy cannot be destroyed, so somehow 1160J of energy was lost by the person. We haven't yet considered friction though, as the...

But then what does that have to do with Z1.

So if you consider that only gravity and friction do work on the person, what can you say. Remember the conservation of energy.

What does that mean rotated about z1 by 450? Could you elaborate?

Okay, so then how much kinetic energy would she have if there was no friction? (v=9.7m/s) How much kinetic energy does she actually end up with? What happened to the energy?

Sure, that is a good first step. How about this, IF there was no friction, how fast would you expect the person to be going at the bottom of the slide?

Use the work energy principle, \Delta E_k=W What forces are doing work on the person?

You are welcome ;).

Yea, that sounds great =-).

That's right ;).