Ok, I think that I finally get it. No garuntee I'll be able to do a similar problem in the future. But thanks for all your help. I really appreciate it. :D
So, that gives us a force on the floor of 7.105 N.
The next part of the question asks about the acceleration of the lizard during the launch time so 7.105/mass=142.1 m/s^2. But doesn't this sound a little too fast?
Ok, so...
KE during contact with floor=.05*9.8*.01=.06615 J
.06615=Force(.01); Force=.49 N during contact with the floor
This gives us a net force of 1.01 N on the lizard.
Since force due to gravity while the lizard is in contact with the floor=mg=.05(9.8)=.49 N.
And assuming...
I'm still lost. Every formula I know for finding velocity over some distance depends on the time it took to travel that distance.
The only solution I can come up with looks like this:
Vfinal=(2gh)^(1/2), where g=9.8 and h=.15-.015=.135
So, Vfinal=1.63 m/s
KE=1/2mv^2, where v=1.63...
I know that I'm going to sound really stupid, but I still don't get it. I skipped this problem and moved on to others, and did what I could with those, but now I'm just more confused than ever when I look back at this one. :S
[SOLVED] force in a non-isolated system
I don't normally do something like this, but I'm stuck on my homework.
A lizard jumps, pushing with its legs against the ground. Assume the lizard's inertia is .05 Kg. Its center of mass starts out .5 cm above the floor, and raises 1 cm to 1.5 cm...