The cable of the 1800 kg elevator cab in the figure snaps when the cab is at res

Question

The cable of the 1800 kg elevator cab in the figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 4.0 m above a spring of spring constant k = 0.33 MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 3.1 kN opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance x that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance (above the point of maximum compression) that the cab will bounce back up the shaft. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)

Explanation / Answer

m = mass of the elevator = 1800 kg
h = initial height = 4 m
k = spring constant = 330000 N/m
R = frictional force = 3100 N
g = acceleration by gravity = 9.8 m/s²

a)
The gravitational potential energy of the elevator (relative to the point of contact with the spring) is:
GPE = m×g×h
GPE = 1800×9.8×4
GPE = 70560 J

The energy 'lost' due to friction while falling a distance h is:
Wr = R×h
Wr = 3100×4
Wr = 12400 J

The remaining energy is the kinetic energy of the elevator, so
KE = GPE - Wr
KE = 70560 - 12400
KE = 58160 J

Kinetic energy is also given by:
KE = mv²/2
so
v = ( 2×KE / m )
v = ( 2×58160 / 1800 )
v = 8.04 m/s

b)
Let X be the distance the spring is compressed.
The TOTAL distance the elevator falls is h + X.
The gravitational potential energy of the elevator (relative to the lowest point) is:
GPE = m×g×(h + X)

The energy 'lost' due to friction wile falling a distance h + X is:
Wr = R×(h + X)

The remaining energy is stored as elastic potential energy in the spring.
Elastic potential energy is given by
EPE = k×X²/2

so
EPE = GPE - Wr
k×X²/2 = m×g×(h + X) - R×(h + X)
k×X²/2 = m×g×h + m×g×X - R×h + R×X
k×X²/2 - m×g×X - R×X = m×g×h - R×h
k×X²/2 - (m×g - R)×X - (m×g + R)×h = 0

Filling in the numbers, you get:
330000×X²/2 - (1800×9.8 - 3100)×X - (1800×9.8 + 3100)×4 = 0
165000×X² - 14540×X - 82960 = 0

Solve this quadratic equation using the discriminant (abc formula)
A negative compression makes no sense in this case, so the only valid solution is
X = 0.755 m

c)
Let H be the distance the elevator bounces back up.
The elastic potential energy stored in the spring at its maximum compression is:
EPE = k×X²/2
EPE = (330000 N/m)×(0.755 m)²/2
EPE = 94054.125 J

The gravitational potential energy of the elevator at height H is:
GPE = m×g×H
GPE = (1800 kg)×(9.8 m/s²)×H
GPE = (17640 N)×H

The energy lost due to friction while going up to height H is:
Wr = R×H
Wr = (3100 N)×H

By conservation of energy, you know that:
EPE = GPE + Wf
so
(94054.125 J) = (17640 N)×H + (3100 N)×H
(94054.125 J) = (20740 N)×H
H = 4.535 m


d)
Let D be the total distance that the cab will move before coming to rest.
From part (c), you know that the elastic potential energy stored in the spring at its maximum compression (after the first fall) is:
EPE = 94054.125 J
All that energy must be 'consumed by friction'.
So
(3100 N)×D = (94054.125 J)
D = 30.34 m

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