Recent content by jmm5180

I am not sure how to deal with curvature except in circles; we haven't covered it yet.

The radius is half the height, so 56m at that speed of 7.03m/s (this was an online simulation and that speed was measured in the simulation).

I'm only in AP Physics 1, so we haven't done any functions yet, but the hill can only be as high as the kinetic energy from the first hill will allow it to be... 0.5mv^2 = mgh V^2 = 2gh r = 2gh/g or 2h, so double the height of the last hill, but that can't be right.

Homework Statement Based on the height of a first hill (114.5m), mathematically determine the minimum height (circular motion and radius) of the next hill based on the speed generated on the first hill. At the bottom of the first hill, the velocity of the roller coaster is 47.38m/s. There is...

Homework Statement A 165N sled is pulled a distance of 205m. The task requires 5200J of work to overcome friction and is done by pulling a rope at what angle? Coefficient of friction is 0.25. Homework Equations W= Fdcos@ (@ for theta) F= ma The Attempt at a Solution My free body diagram had...

Yes, thank you!

Homework Statement How much work is done by friction when a box with an apparent mass of 325kg moves horizontally across a floor with coefficient of friction of .33 and is pulled at 59.9 degree angle for 2.60 m? What is the force that is pulling the box? Homework Equations W=Fdcos@ (@= theta)...

Thank you. I've used -9.8 m/s^2 before in questions and have gotten the correct answer. How do I know when to use the positive/negative version?

Homework Statement Calculate the force required to push a 22kg sled down a 6 degree hill and have an ending speed of 60km/hr after traveling 75m when the coefficient of friction is 0.20. Homework Equations Fnet= ma (V2)^2 = (V1)^2 + 2a(d2-d1) Ff= uN W= mg The Attempt at a Solution I'll use @...

Interesting, thank you. I didn't realize that they would lead to different results.

It was right! Thank you SO much!

Flat road: Ff=ma N=W=mg uN = ma umg = ma u = a/g Hill: N = Wy N= wcos(theta) Ff= ma - wsin(theta) uN = ma - wsin(theta) uwcos(theta) = ma - wsin(theta) Umgcos(theta) = ma - mgsin(theta) u= (ma - mgsin(theta)) / mgcos(theta) U= (a - gsin(theta)) / gcos(theta) (a1/g) = (a2- gsin(theta)) /...

The positive x-axis is pointing up the hill, and the negative x-axis is pointing down the hill. The motorcycle is at the origin. The normal force lies on the positive y axis.

How do I define my axes?

It is unclear to me what cases you refer to and which direction you have designated as y. Please keep in mind that I cannot see your drawing. By first scenario I mean when the motorcycle is traveling on a flat road. And second scenario is up the hill. The normal force is along the Y axis and...