The brothers of Iota Eta Pi fraternity build a platform, supported at all four c

Question

The brothers of Iota Eta Pi fraternity build a platform, supported at all four corners by vertical springs, in the basement of their frat house. A brave fraternity brother wearing a football helmet stands in the middle of the platform; his weight compresses the springs by a distance 0.18m. Then four of his fraternity brothers, pushing down at the corners of the platform, compress the springs another distance 0.56m until the top of the brave brother's helmet is a distance 0.86m below the basement ceiling. They then simultaneously release the platform. You can ignore the masses of the springs and platform. A) When the dust clears, the fraternity asks you to calculate their fraternity brother's speed just before his helmet hit the flimsy ceiling. B) Without the ceiling, how high would he have gone?

Explanation / Answer

This is an exercise inspring constantsand energy transfer.

The first distance given allows you to calculate the spring constant of the construction. (it will be in terms of the brother's mass; don't worry, it factors out later.)

k = m * g / x where x = 0.18m

When the rest of the brothers pull the platform lower, you can calculate the stored energy.

Us = 0.5 * k * x^2
Us = 0.5 * m * g * x where x = 0.71 m

When the brothers release the platform, the stored energy is converted to kinetic energy and gravitational potential energy of the brave fraternity brother (BFB) until the springs reachequilibrium position.
PEgrav = mgh where h = 0.71 m
KE = 0.5*mv^2

At equilibrium position, the energy originally stored in the spring will be equal to the energy of the BFB. 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, mass factors 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 will show that from the equilibrium point, the BFB has 0.19 meters between his helmet and the ceiling. Calculate his speed after he has traveled this distance.

v = v0 + At
x = x0 + v0t + 0.5At^2

Solve 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 smaller value is the ceiling crossing on the way up, and the larger is on the way back down, assuming no ceiling is actually present.


1.
With t, solve for v:
v = 2.639 m/s + (-9.81 m/s^2) * 0.090 s
v = 1.756 m/s

2.
with initial v, solve for x:
v = v0 + At
t = (0 - 2.639 m/s) / -9.81 m/s^2
t = 0.2690 seconds

x = x0 + v0t + 0.5*At^2
x = 0 + (2.639 m/s * 0.2690 s) + (0.5 * -9.81 m/s^2 * 0.2690^2 s^2)
x = 0 + .71 m - .3549 m
x = 0.3551 meters (above the platform)

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