In a \"worst-case\" design scenario, a 2000-kgelevator with broken cables is fal

Question

In a "worst-case" design scenario, a 2000-kgelevator with broken cables is falling at 4.00 m/swhen it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 17000-N frictional force to the elevator.

A.What is the speed of the elevator after it has moved downward 1.00 m

from the point where it first contacts a spring?

B.When the elevator is 1.00

m below point where it first contacts a spring, what is its acceleration?

Explanation / Answer

Mass , m= 2000 kg

speed , u = 4 m/s

let the spring cosntant is k

Now , for spring to stop the car in 2 m

Using conservation of energy

0.5 * k * 2^2 + 17000 * 2 = 2000 * 9.8 * 2 + 0.5 * 2000 * 4^2

solving for k

k = 10600 N/m

a)

let the speed of elevator is v m/s

for x = 1 m

Using conservation of energy

0.5 * k * x^2 + F * x = m * g * x + 0.5 * m * u^2

0.5 * 10600 * 1^2 + 17000 * 1 + 0.5 * 2000 * u^2 = 2000 * 9.8 * 1 + 0.5 * 2000 * 2^2

solving for u

u = 1.14 m/s

the speed of elevator will be 1.14 m/s

b)

let the acceleration is a

using second eqation of motion

m *a = k *x + F - mg

2000 * a = 10600 * 1 + 17000 - 2000 * 9.8

solvnig for a

a = 4 m/s^2

the acceleration at this instant is 4 m/s^2

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