lax400 Posted April 8, 2008 Posted April 8, 2008 A boulder with a mass of 2,500kg on a ledge 200m above the ground falls. What is the speed of the boulder just before it hits the ground. To find velocity you need to use the kinetic energy formula right? But when I tried to plug in everything, it wouldn't work. IT only looks as if the GPE formula would work... could someone help me please?
Klaynos Posted April 8, 2008 Posted April 8, 2008 You might be able to do it with just the kinematic equations, you know: u,a,s, and want to find v....so you have/want 4 of the 5 variables so yep the kinematic equations can be used here. Do you know what they are?
ydoaPs Posted April 8, 2008 Posted April 8, 2008 When it is motionless, it is at it's highest point. Here, the total mechanical energy is solely composed of potential energy. It's fastest point is when it is at height=0 and thus the total mechanical energy is solely composed of kinetic energy. If we disregard air friction, we can say that all of the potential energy was converted to kinetic energy in the fall. You can then set the initial potential energy equal to the final kinetic energy. Then you can solve for the velocity.
lax400 Posted April 9, 2008 Author Posted April 9, 2008 You might be able to do it with just the kinematic equations, you know: u,a,s, and want to find v....so you have/want 4 of the 5 variables so yep the kinematic equations can be used here. Do you know what they are? umm no sorry im only in 9th grade =\
ydoaPs Posted April 9, 2008 Posted April 9, 2008 Kinematics....It's usually taught first. It's the equations of motion. vf=v0+at xf=x0+vt+(0.5)at2 etc.
ydoaPs Posted April 9, 2008 Posted April 9, 2008 Ok, you had some idea of what I said the first time. Go with that.
Cap'n Refsmmat Posted April 9, 2008 Posted April 9, 2008 You can use kinetic and potential energy. [math]PE = mgh[/math] where m = mass, g = the acceleration due to gravity (9.8m/s2) and h = the height. All of that energy will be converted to kinetic energy by the time the object hits the ground. [math]KE = \frac{1}{2} mv^2[/math] where m = mass and v = velocity. So make the kinetic energy equal the potential energy you found in the first step and solve for velocity.
Klaynos Posted April 9, 2008 Posted April 9, 2008 You can do it with energies, just equate the gravitational energy to kinetic and solve for v.
ydoaPs Posted April 9, 2008 Posted April 9, 2008 uhh.. how do i do that... It's what I and Cap'n said to do.
thedarkshade Posted April 9, 2008 Posted April 9, 2008 What is the speed of the boulder just before it hits the ground.You can try with [math]v=\sqrt{2gh}[/math].
Klaynos Posted April 9, 2008 Posted April 9, 2008 You can try with [math]v=\sqrt{2gh}[/math]. Which can be derived either using kinematics or energy...
5614 Posted April 9, 2008 Posted April 9, 2008 The potential energy it gains by dropping 200m is found by using: PE = mgh You know the mass, g is acceleration due to gravity (you'll use 10, or 9.8 or 9.81) and h is the distance it falls. Then you know that all of that potential energy went in to kinetic energy, and you know the formula: KE = ½mv². So you say that all the PE goes to KE, therefore: PE = KE mgh = ½mv² gh = ½v² 2gh = v² v = √(2gh) That should help. You understand now?
Hadron Posted April 9, 2008 Posted April 9, 2008 A boulder with a mass of 2,500kg on a ledge 200m above the ground falls. What is the speed of the boulder just before it hits the ground. To find velocity you need to use the kinetic energy formula right? But when I tried to plug in everything, it wouldn't work. IT only looks as if the GPE formula would work... could someone help me please? Remember that mass does not at all enter into solving this problem.
ydoaPs Posted April 9, 2008 Posted April 9, 2008 Remember that mass does not at all enter into solving this problem. For this problem. It might not always be the case. What if we add in a electric potential into the mix?
Cyclic Posted April 10, 2008 Posted April 10, 2008 this specific equation (of motion) does not need mass. use equation: 2 2 Vf = Vi + 2a (x) Vf = ? Vi = 0 a = 9.8m.s x = 200m Vf should = 62.61m.s downwards Does this help?
thedarkshade Posted April 10, 2008 Posted April 10, 2008 this specific equation (of motion) does not need mass. Equations of motions don't really need mass on them, unless you want to find energy out of them.
ydoaPs Posted April 10, 2008 Posted April 10, 2008 Equations of motions don't really need mass on them, unless you want to find energy out of them. What if your force isn't gravity at sea level?
thedarkshade Posted April 10, 2008 Posted April 10, 2008 What if your force isn't gravity at sea level? Perhaps a little misunderstanding here but I was referring to the equations of velocity!
ydoaPs Posted April 10, 2008 Posted April 10, 2008 Perhaps a little misunderstanding here but I was referring to the equations of velocity! What do you think causes the "a"? The only reason you don't need mass is because we know the acceleration at sea level due to gravity.
Klaynos Posted April 10, 2008 Posted April 10, 2008 Also you must consider situations where the mass changes...
ydoaPs Posted April 10, 2008 Posted April 10, 2008 Also you must consider situations where the mass changes... I guess thedarkshade isn't a rocket scientist.
timo Posted April 10, 2008 Posted April 10, 2008 A simple e.o.m including mass would be that for an electrically charged object with charge q in an electric field E: [math]\ddot x = qE/m[/math].
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