Recent content by saturdaykate

Thank God! And thank you for your help! I won't even ask you for help on the third part, I think I'll just leave it blank, but thanks for your help...

actually I just took the square root of 2*g*h (rad2(9.8)(6)) and got 10.8444m/s. This is because, like you were trying to tell me I think, that it is not changing in energy? It is conservative, right?

Okay, so I think I understand question 2. Question one: I did not lose any energy. So... that would mean that my energy at the beginning of the slide woudl equal that at the end of the slide? So would I set potential energy and kinetic energy equal to each other?

so, is using the U equation ( I think my teacher called it the "energy final-energy initial" formula?) going to find me the initial velocity? I AM SO LOST.

All the information that I have on the problem was given to you in the header. If you are asking me, I would guess that it lost energy to friction?

oh wait, work is equal to force times displacement, right? so would i just set that whole mess equal to (2kg)(9.8m/s2)(6m) --- (the potential energy)?

Is the work equal to the kinetic energy added to the potential energy?

Well, for part 1, I thought that I would use (MGHf+(1/2)MVf^2)-(MGHi+(1/20MFi^2) Then, assuming the block starts at rest, you would get ((2kg)(9.8m/s2)(0m)+(1/2)(2kg)(Vf^2))-((2kg)(9.8m/s2)(6m)+(1/2)(2kg)(0m/s)) But what do I set that equal to? I am just completely lost. My teacher is...

I have been trying at this stupid problem all day. It was on a test I took, but my whole class couldn't figure out any of the extended response so he let us take it home. For this problem, I kept messing aroudn with the potential energy (mgh) and kinetic energy ((1/2)mv^2) formula but I keep...

thank you, i really appreciate your help.

A 2kg block at the top of an inclined plane at a height h=6m above the ground moves down the plane. 1) Find the value of the velocity when the block is at the bottom of the plane assuming no friction. 2) With friction present, the final velocity is only 7.0 m/x at the bottom of the...

Oh, wait, I think I might have gotten the answer, does 3880.8J seem about right?

I am stuck on this problem and I really need to figure it out or I will probably have to kill myself. Just saying. Also, I'm having the worst day of life, was fired from my job, and basically laughed at by my parents. Here is the problem: A 60kg person starts from rest and slides down the...