Raider5678 Posted April 25, 2017 Posted April 25, 2017 What is the physics formula for an object going uphill? Assuming you know the initial velocity, the mass, the angle of the hill, and there is no friction or air resistance.
Velocity_Boy Posted April 25, 2017 Posted April 25, 2017 What is the physics formula for an object going uphill? Assuming you know the initial velocity, the mass, the angle of the hill, and there is no friction or air resistance. Hey Raider. So.....The equation you're seeking is one that will deal with what's called Frictionless Mass on Incline. Here's a cool page that shows that equation and has accompanying graphics. Now the bad news....A bit of Trig is involved. But only re sin. So not too bad. LOL http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html
Raider5678 Posted April 25, 2017 Author Posted April 25, 2017 (edited) Hey Raider. So.....The equation you're seeking is one that will deal with what's called Frictionless Mass on Incline. Here's a cool page that shows that equation and has accompanying graphics. Now the bad news....A bit of Trig is involved. But only re sin. So not too bad. LOL http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html I want it going up hill. I have already calculated it going down hill. I am perfectly capable of doing trigonometry. Edited April 25, 2017 by Raider5678
Bender Posted April 25, 2017 Posted April 25, 2017 Conservation of energy: velocity only depends on the height. [math]mgh_1+\frac{mv_1^2}{2}=mgh_2+\frac{mv_2^2}{2}[/math] Or dropping the mass: [math]gh_1+\frac{v_1^2}{2}=gh_2+\frac{v_2^2}{2}[/math] 1
Raider5678 Posted April 25, 2017 Author Posted April 25, 2017 Conservation of energy: velocity only depends on the height. [math]mgh_1+\frac{mv_1^2}{2}=mgh_2+\frac{mv_2^2}{2}[/math] Or dropping the mass: [math]gh_1+\frac{v_1^2}{2}=gh_2+\frac{v_2^2}{2}[/math] So distance does not matter? Nor angle?
Velocity_Boy Posted April 26, 2017 Posted April 26, 2017 I want it going up hill. I have already calculated it going down hill. I am perfectly capable of doing trigonometry. It's good to know you're capable. My link did offer an ascending grade equation. Or what you do quaintly referred to as uphill. You just had to, you know, scroll down a bit and not dismiss it outright because the first graphic depicted a descending grade equation. Or, to you, downhill.
Raider5678 Posted April 26, 2017 Author Posted April 26, 2017 It's good to know you're capable. My link did offer an ascending grade equation. Or what you do quaintly referred to as uphill. You just had to, you know, scroll down a bit and not dismiss it outright because the first graphic depicted a descending grade equation. Or, to you, downhill. No, all equations were for going down the incline.
Bender Posted April 26, 2017 Posted April 26, 2017 (edited) So distance does not matter? Nor angle? Not if there is no friction or other outside forces and the velocity is the only thing you want to know. This can vastly simplify otherwise complex problems, such as roller coasters with loopings. Edited April 26, 2017 by Bender
Sriman Dutta Posted April 26, 2017 Posted April 26, 2017 (edited) I derived the equation here: [math] KE_{total} = Work_{gravity} + Work_{upward} [/math] [math] \frac{1}{2}mv^2 = mgh. + mah cosec \theta[/math] [math] \frac{v^2}{2}=h(g+a cosec \theta)[/math] a is the acceleration of the body up the slipping surface. Edited April 26, 2017 by Sriman Dutta
swansont Posted April 26, 2017 Posted April 26, 2017 F = ma, and the mathematical definitions of the terms in kinematics You can derive what you need from that.
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