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Nata [24]
3 years ago
8

In a "worst-case" design scenario, a 2000-kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushion

ing 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. 1) What is the speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring? 2) When the elevator is 1.00 m below point where it first contacts a spring, what is its acceleration?
Physics
1 answer:
Fofino [41]3 years ago
3 0

Answer:

Part a)

v_f = 3.65 m/s

Part b)

a = - 4 m/s^2

Explanation:

Part a)

As we know that the elevator is stopped while spring is compressed by x = 2 m

now we will have

\frac{1}{2}kx^2 + F_f x - mgx = \frac{1}{2}mv^2

\frac{1}{2}k(2^2) + (17000)(2) - (2000\times 9.81\times 2) = \frac{1}{2}(2000)(4^2)

2k + 34000 - 39240 = 16000

k = 10620 N/m

Now we have to find the speed when spring is compressed by x = 1

So again by work energy theorem

W_g + W_f + W_{spring} = \frac{1}{2}m(v_f^2 - v_i^2)

2000(9.81)(1) - 17000(1) - \frac{1}{2}(10620)(1^2) = \frac{1}{2}(2000)(v_f^2 - 4^2)

-2.69 = v_f^2 - 16

v_f = 3.65 m/s

Part b)

Net force on the elevator while spring is compressed by x = 1

F_{net} = mg - kx - F_f

F_{net} = 2000(9.81) - (10620\times 1) - 17000

F_{net} = -8000 N

now acceleration is given as

a = \frac{F}{m}

a = -\frac{8000}{2000}

a = - 4 m/s^2

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