Fraternity Physics.?

Question

The brothers of a fraternity build a platform in their basement, supported at all four corners by vertical springs. A brother wearing a helmet stands in middle of the platform; his weight compresses springs by a distance 0.200 . Then four of his brothers, pushing down at corners of the platform, compress springs another distance 0.510 until top of brother's helmet is distance 0.920 below the basement ceiling. They then simultaneously release the platform. You can ignore masses of the springs and platform.

A:When the dust clears, the fraternity asks you to calculate their brother's speed just before his helmet hit the flimsy ceiling.

B:Without the ceiling, how high would he have gone?

C:In discussing their probation, the dean of students suggests that the next time they try this, they do it outdoors on another planet. Would the answer to part (B) be the same if this stunt were performed on a planet with a different value of g? Assume that the fraternity brothers push the platform down 0.510 as before.

Answer 1

The energy stored in the springs when the brother first stands on the platform is
equal to the change in potential energy of the brother over the compression distance

m*g*.2=.5*k*.2^2
Note that the fact that there are four springs is just absorbed by the k
The four brothers push down on the corners which adds additional energy stored in the springs, and an additional loss in potential energy. The energy in the springs is
.5*k*.710^2

at a distance of .920 the kinetic energy of the helmeted brother is

.5*m*v^2

which is equal to the recovery of the energy in the springs and the loss of potential energy to fly upward .920

.5*m*v^2=.5*k*.710^2-m*g*.920

v^2=k*.710^2/m-2*g*.920

from the initial measurement of the brother standing on the platform, we can solve for k/m
k/m=2*g*.2/.2^2

plug into the equation for v^2
v^2=2*g*.2*(.710/.2)^2-2*g*.920

v=sqrt(v^2)

solving
v=5.6 m/s

The height he would have flown would be the conversion of the energy in the springs to potential energy at apogee

.5*k*.710^2=m*g*h
h=.5*k*.710^2/(m*g)

again, k/m =2*g*.2/.2^2

h=.2*(.710/.2)^2

h=2.52 m

The distance flown would be different since the initial compression distance that we used to express k/m would be different

recall
m*g*.2=.5*k*.2^2

on another planet, this would be

m*g'*d=.5*k*d^2
Where d is the distance of compression, and where g' is gravitational constant on the other planet
k/m=2*g'/d

the height to apogee would be

h=(.510+d)^2/d

let's see if we can express d in therms of g and g',

the original k/m=2*g/.2
the new k/m = 2*g'/d
k/m is constant, so
d/.2=g'/g
or
d=.2*g'/g

plugging into the height equation:

h=(.510+.2*g'/g)^2/(.2*g'/g)

For g'=g, we get the Earth version (a quick check of my algebra)

For g'>g, the h decreases, although slowly since the four other brothers compress the springs an additional .510 from the compression distance due to the weight of the helmeted one.

for example, for g'=2*g

h=(.510+.4)^2/.4
=2.1

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