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An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should happen to break at a height h above the top of the spring calculate the value that the spring constant k should have so that passengers undergo an acceleration of no more than 5.0g(5*9.8) when brought to rest. Let M be the total mass of the elevator and passengers.
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The answer behind the book is:
12Mg/h
I'm really stuck. I have over 2 hours with this problem. PLEASE help. THx in advance.

2007-03-20 09:12:19 · 3 answers · asked by Anonymous in Science & Mathematics Physics

3 answers

Acceleration is variable here, so uniformly accelerated motion formulas are not applicable. Maximum acceleration occurs at the lowest point in trajectory, when the elevator car stops momentarily before reversing the direction of motion. At this point, acceleration should not exceed 5.0 g. Two forces act upon the car:

1. Its weight, W.
2. Spring reaction, F (upward).

Thus,
Ma = F − W = kY − Mg,

where Y stands for maximum spring compression (spring length change); but a ≤ 5 g, so,
kY − Mg ≤ 5 Mg, or
kY ≤ 6 Mg. Then,
Y ≤ 6 Mg/k.

Maximum admissible value for Y is then 6 Mg/k.

Y = 6 Mg/k..............(1)

Distance between car and spring is h. Car falls this much, and an extra Y m beyond that point. Total change in potential energy is Mg(h + Y). This is converted to elastic potential energy in the spring. Thus,

½ kY² = Mg(h + Y)........(2)

Substituting (1) in (2),

½ k(36 M²g²/k²) = Mg(h + 6 Mg/k)
18 Mg/k = h + 6 Mh/k
12 Mg/k = h
k = 12 Mg/h.

2007-03-20 21:34:54 · answer #1 · answered by Jicotillo 6 · 0 1

The maximum deceleration occurs when the force of the spring is at a maximum, which is the instant before the elevator comes to rest.

Using conservation of energy, the total potential energy of the elevator and passengers is
M*g*(h+x)
where x is the compression distance of the spring

the energy will be converted to energy stored in the spring, which is
.5*k*x^2

when compressed the the force is required to be
M*g*5
and the force of the spring is
k*x

so
x=M*g*5/k
and
x^2=(M*g*5/k)^2
from above

.5*k*x^2=
M*g*(h+x)

multiply both sides by 2
k*x^2=2*M*g*(h+x)
plugging in the x and x^2
25*M^2*g^2/k=
2*M*g*h+
10*M^2*g^2/k

simplify
M*g*15/k=2*h
k=M*g*15/(2*h)

is that what you got?

j

2007-03-20 10:11:47 · answer #2 · answered by odu83 7 · 0 1

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2016-09-05 09:41:27 · answer #3 · answered by ? 4 · 0 0

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