Recent content by _h2tm

OK, thanks. I completely get how to do this in the future. Regarding your second comment though -- I understand about arctan. I drew a diagram so I knew where my vectors were. Was this comment just FYI or was my attempted method something that would work, I was just simply doing it incorrectly?

Homework Statement Find the power input of a force \vec{F} acting on a particle that moves with a velocity \vec{V} for each of the following situations. \vec{F} = 4\hat{i} N  + 3\hat{j} N , \vec{V} = 7\hat{i} m/s  \vec{F} = 7\hat{i} N  - 5\hat{j} N , \vec{V} = -5\hat{i} m/s...

Maybe try looking up the average speed of an elevator and take half of that ... or just go with .01 m/s.

I may not be understanding your question, but it seems like you should start with the Pythagorean Identities.

What I wrote above included | & | to indicate taking the absolute value and gave the correct result (at least the answer that the website was looking for). I'm not seeing how I can manipulate the sum of the forces so it would be R(g - .35g).

After getting more information from my teacher, here is the solution for the "smallest" value, i.e. the speed at the bottom of the roller coaster: Assume Fsp = .35mg at the bottom of the roller coaster as well the top. Fsp - mg = mv2/R .35g - g = v2/R √|R(.35g - g)| = v The question...

When I entered the answer (we use WebAssign for our homework), it showed as correct in the space that said "largest value" :).

Yes. I copied & pasted the problem: The "largest" value is the one I found above (≈13.62 m/s).

We haven't covered conservation of energy yet (we've just finished Newton's laws, friction, uniform circular motion, drag forces, and an intro to work and dot-products). I sum the forces at the bottom of the roller coaster: ƩF = Fsp - mg = mv2/R ∴ Fsp = m[(v2/R) + g] I don't know the...

Homework Statement The radius of curvature of a loop-the-loop for a roller coaster is 14.0 m. At the top of the loop, the force that the seat exerts on a passenger of mass m is 0.35mg. Find the speed of the roller coaster at the top of the loop. (Enter your answers from smallest to largest.)...