1) An elevator cable breaks when a 950 kg elevator is 20.5 m above the top of a huge spring ( k= 6.00×10^4 N/m) at the bottom of the shaft.
A) Calculate the amount the spring compresses (note that here work is done by both the spring and gravity).
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Approach:
When the spring is compressed a distance x, it is storing potential energy PEs given by the equation (see Source 1):
(1) PEs = ½ * k * x²
The source of this energy is the work that gravity does Wg on the elevator in pulling it down the distance of free-fall y plus the distance of compression x.
(Note that this is the same as the gravitational PE lost by the elevator in falling the distance y + x.)
Equations and calculations:
(2) Wg = m * g * (y + x)
Equating (1) to (2):
(3) ½ * k * x² = m * g * (y + x)
Rearranging:
(4) (½ * k) * x² - (m * g) * x - (m * g * y) = 0
Substituting values for k, m, g, and y we get:
(5) 3.00e4 * x² - (950 * 9.81) * x - (950 * 9.81 * 20.5) = 0 ==>
(6) 3.00e4 * x² - 9.32e3 * x - 1.91e5 = 0
Solving for the roots of this equation (see Source 2), we get:
x1 = 2.68
x2 = -2.37
So, the amount of compression of the spring = 2.68m <<<=== Answer
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