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Chapter 2: Problem 52

The single cable supporting an unoccupied construction elevator breaks when the elevator is at rest at the top of a \(120-\mathrm{m}\) high building. ( \(a\) ) With what speed does the elevator strike the ground? \((b)\) For how long was it falling? \((c)\) What was its speed when it passed the halfway point on the way down? ( \(d\) ) For how long was it falling when it passed the halfway point?

Short Answer

Expert verified
The elevator will strike the ground at a speed of approximately \(49.0 m/s\), after falling for roughly \(4.9 s\). It will pass the halfway point (\(60m\)) at a speed of approximately \(34.6 m/s\), and it will take around \(3.5 s\) to get to this point.

Step by step solution

01

Calculation of the final speed at the ground

To determine the speed at which the elevator hits the ground, the kinematic equation, \(v^2 = u^2 + 2gs\) can be used. Where \(v\) is the final velocity, \(u\) is the initial velocity (which is 0 as the elevator starts from rest), \(g\) is the acceleration due to gravity (approximately \(9.8 m/s^2\)), and \(s\) is the distance fallen (which is the height of the building, \(120 m\)). Substituting these values into the equation and solving for \(v\), we get \(v = \sqrt{2gs}\).
02

Calculation of the time of fall

To calculate the time of fall to the ground, the kinematic equation, \(s = ut + 0.5gt^2\) can be used. Where \(s\) is the distance fallen, \(u\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time of fall. Substituting the known values and solving for \(t\), we can use the square root form as \(t = \sqrt{2s/g}\).
03

Calculation of the speed at the halfway point

To calculate the speed at which the elevator passes the halfway point, we can use the same method as in step 1, but with the distance \(s\) as half the height of the building (i.e. \(60 m\)). This gives us \(v_{half} = \sqrt{2gs_{half}}\), which we can compute.
04

Calculation of the time of fall to the halfway point

Lastly, to figure out the time it takes to fall to the halfway point, we can apply the same method as in step 2 but with the distance \(s\) as half the height of the building. This yields \(t_{half} = \sqrt{2s_{half}/g}\) which can be calculated to achieve the specific time.

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Most popular questions from this chapter

A dog sees a flowerpot sail up and then back down past a window \(1.1 \mathrm{~m}\) high. If the total time the pot is in sight is \(0.54 \mathrm{~s}\), find the height above the top of the window to which the pot rises.

A rock is dropped from a 100 -m-high cliff. How long does it take to fall \((a)\) the first \(50.0 \mathrm{~m}\) and \((b)\) the second \(50.0 \mathrm{~m}\) ?

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