Recent content by septum

I... finally came to the answer after about 8 pieces of paper. I think far too into this. Thank you for your help

I still don't get this at all.

Homework Statement Suppose a slide similar to Der Stuka is 36.0 meters high, but is a straight slope, inclined at 45° with respect to the horizontal. (a) Find the speed of a 51.1 kg thrill seeker at the bottom of the slide, assuming no friction. 26.5 m/s (b) If the thrill seeker has a speed...

I can't picture anything being pulled up at an angle without being on an incline so I'm probably not the person to be trying to help. :/ Sorry

Okay, well as I understand it for part A normal force is equal to mgcos50 degrees. I don't think you have to multiply anything but gravity by your mass. Thus, n=mgcos: n=(80kg)(9.8 m/s^2)*cos(50)

*facepalm* Thank you so much for your help.

part B would have to be sin(-70) since it is below, correct?

Even if I do 2.00m*cos25 degress it comes out as 1.81 m. If I plug that into my 140 N * (1.81 m) = 153 which is far too large. It must be much less and that is what I'm failing to understand.

T.T i don't understand this at all. I drew a whole lot of right triangles solving for a h and o using 90 degrees and 25 degrees and 65 degrees (basically everything I can think of) and get small numbers such as 2.21 and when I add that in that gives you a.. LOT of joules which is nonsensical.

Homework Statement A 140-N child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy associated with the child relative to her lowest position at the following times. (a) when the ropes are horizontal I get this one. I used PE=mgy...