Recent content by myelevatorbeat

I am currently working on a lab report for my physics class. During the lab, we used air tracks, gliders, and a photogate to measure the value of 'g'. Basically, we would raise one end of the air track to a certain height and let the glider slide down the frictionless track and the timer would...

V^2=Vo^2-2g(Y-Yo), sorry.

Still working on figuring this problem out. So far, I figured out the Potential Energy of the spring and it is 260 Joules. Now, I used this equation: PEo+KEo=PE+KE (with KEo=0 because the box isn't in motion) So: PEo=PE+KE 260 J = mgH+1/2mv^2 260 J = (0.300 kg)(9.80 m/s^2)H+1/2(0.300 kg)v^2...

Ok so I did this: 260 J =(5.20 x 10^3 kg)(9.80 m/s/s)h and solving for h I got: h=0.005 m However, when I put this answer into webassign.net it tells me I'm incorrect because of "orders of magnitude". Did I do something wrong?

Wait, I meant PE and I guess the spring has a PEo=260 J and the box has a PEo=0 since it is at rest.

I know that PEo+KEo=PE+KE also. I would assume I'd need two equations since since the KEo of the spring is 260 Joules and the KEo of the box is 0?

So now, would I just plug 260 J into: mgh to figure out h or is it more complicated than that?

I would assume the energy at the top would be 260 J since that was the potential energy of the spring and the box had no potential energy since it was sitting at rest?

If I solve for a in F=ma, I get: -520 N = (5.20x10^3 kg)a a= -0.100 I don't think this is right because I'd expect an object sitting on a spring to accelerate positively, until the end when it would become negative and begin slowing down?

Homework Statement A 0.300 kg block is placed on a light vertical spring (k = 5.20 103 N/m) and pushed downward, compressing the spring 0.100 m. After the block is released, it leaves the spring and continues to travel upward. What height above the point of release will the block reach if...

Homework Statement A daredevil on a motorcycle leaves the end of a ramp with a speed of 39.0 m/s as in Figure P5.23. If his speed is 36.8 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance. The Attempt at a Solution...

Let me try to explain where I'm not getting it: I understand now where the second cosx2 came from, but I wind up with this equation: T1+T1=Fgcosx2/sinx1cosx1sinx2cosx2 Doesn't that mean it's really 2(T1)=Fgcosx2/sinx1cosx1sinx2cosx2 I want to know how to get just...

Homework Statement Find the acceleration reached by each of the two objects shown in Figure P4.49 if the coefficient of kinetic friction between the 7.00kg object and the plane is 0.250 Here is a picture of the problem...

So, should I have: T1sinx1cosx2+T1cosx1sinx2cosx2=Fgcosx2? I still don't understand what happens to the second T1

Here's a diagram: http://a373.ac-images.myspacecdn.com/images01/113/l_db80f0d296b290d3e3b1ef076f5fe74c.jpg Sorry about that.