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 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. 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?

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aristocles aristocles · Answer 1 · 2019-10-25T23:08:21+02:00

Answer:

Part a)

[tex]v_f = 3.65 m/s[/tex]

Part b)

[tex]a = - 4 m/s^2[/tex]

Explanation:

Part a)

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

now we will have

[tex]\frac{1}{2}kx^2 + F_f x - mgx = \frac{1}{2}mv^2[/tex]

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

[tex]2k + 34000 - 39240 = 16000[/tex]

[tex]k = 10620 N/m[/tex]

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

So again by work energy theorem

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

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

[tex]-2.69 = v_f^2 - 16[/tex]

[tex]v_f = 3.65 m/s[/tex]

Part b)

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

[tex]F_{net} = mg - kx - F_f[/tex]

[tex]F_{net} = 2000(9.81) - (10620\times 1) - 17000[/tex]

[tex]F_{net} = -8000 N[/tex]

now acceleration is given as

[tex]a = \frac{F}{m}[/tex]

[tex]a = -\frac{8000}{2000}[/tex]

[tex]a = - 4 m/s^2[/tex]