Question

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?

Answer 1

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

$a = \frac{F}{m}$

$a = -\frac{8000}{2000}$

$a = - 4 m/s^2$