**please help: velocity question (only distances given)? - questions to ask fraternity brothers
I have many methods, but none of them gave me the right answer. Can anyone help? I know the right answer for part 1 (3.1) and 2 (1.4), but do not know how?
The brothers of the Eta Iota Pi fraternity to build a platform with four corners of the vertical columns in the basement of his fraternity supported. A brother, fraternity wore a football helmet is in the middle of the platform to be good weight compresses the spring a distance of 0.18 m. Then four of his brothers, brotherhood, and press the edges of the platform to compress other sources of 0.53 m spacing over the helmet of the valiant brother is a distance of 0.90 m below the ceiling of the cellar. Then, while the platform. Can you ignore the masses, orand docks and platforms.
1. As approved by the dust, "said the Brotherhood in order to calculate the speed of your brother, brotherhood, shortly before his helmet struck the roof of fragility?
2. Without the roof gone, this could be?
Thank you.
Sunday, February 21, 2010
Questions To Ask Fraternity Brothers **please Help: Velocity Question (only Distances Given)?
2 comments:
It is a constant exercise in the spring and energy transfer.
The first series of remote control allows you to calculate the spring constant construction. (will be in relation to the mass of a brother, do not worry, the factors that later).
k = m * g / x, where x = 0.18 m
If the rest of the brothers take the lower platform, you can calculate the stored energy.
* We * k = 0.5 x ^ 2
Us = 0.5 * m * g * x, where x = 0.71 m
When the brothers for the platform to let go, the stored energy is converted into kinetic energy and gravitational potential energy of the brother of the brave brotherhood (BFB) that enter the port of the equilibrium position.
PEgrav = mgh, where h = 0.71 m
KE = 0.5 * mv ^ 2
In the rest position, the energy is initially stored in the spring equals the energy of the CBA. Solve for v.
0.5 * M * (-9.81 m / s ^ 2) * 0.71 m = m * (-9.81 m / s ^ 2) * 0.71 m + 0.5 * m * v ^ 2
As promised, are the factors of the mass out.
0.5 * (-9.81 m / s ^ 2) * 0.71 m = (-9.81 m / s ^ 2) * 0.71 m + 0.5 * v ^ 2
-0.5 * (-9.81 M / s ^ 2) *0.71 m = 0.5 * v ^ 2
9.81 m / s ^ 2 * 0.71 m = v ^ 2
v = sqrt (6.9651 m ^ 2 / s ^ 2)
v = 2.6391 m / s
A little arithmetic shows that the equilibrium point of the ABC, 0.19 meters between the hull and the ceiling. Calculate the speed after traveling this distance.
v = V0 + A
x = X0 + V0T + 0.5At ^ 2
Solving for T:
0.19 m = 0 m + 2.6391 m / s * t + 0.5 * -9.81 m / s ^ 2 * t ^ 2
You'll find that t = 0.090 s and 0.357 s. The lower value of the roof across the street, and is driven largely by the way, if no limit is really there.
1.
To be solved with T V:
v = 2.639 m / s + (-9.81 m / s ^ 2) * 0,090 s
v = 1.756 m / s
2.
v with an initial solution for x:
v = V0 + A
T = (0 to 2639 m / s) / -9.81 m / s ^ 2
t = 0.2690 seconds
x = X0 + V0T + 0.5 * A ^ 2
x = 0 + (2,639 m / s * 0.2690 s) + (0.5 * -9.81 m / s ^ 2 * 0.2690 ^ 2 s ^ 2)
= 0 x + .71 m - 3549 m
x = 0.3551 m (above the platform)
Because of his commitment four springs 53 m, the Frat brothers give a potential energy e_0 = (.53 m) * (4) * k on the platform, where k is our constant source and m is the mass of our fellow brethren.
The platform is to press the spoiled frat boy, when's your head at a distance from 9 to 53 = 37 m from the ceiling. The energy needed to lift his head against the ceiling, from this point, we call E_1 = 37 * M * g.
Our energy point of contact is e_f = e_0 - E_1 = 4k (.53) - (.37) m * g. This is the same as its kinetic energy at this point.
Accordingly, 1 / 2 * m * v ^ 2 = 4k (.53) - (.37) m * g. The solution for V and its speed, when his head on the ceiling.
Without the roof, which simply means our e_0 m * g * h, because the potential only work against gravity, spoiled frat boys to be moved in the air.
(.53 M) * (4) * k = m * g * h. Clear H has its maximum height.
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