An elevator accelerates upward at 1 m/s2. When the elevator is moving upward at

Question

An elevator accelerates upward at 1 m/s2. When the elevator is moving upward at 2 m/s, a bolt drops from the ceiling which is 2.5 m above the ?oor. Assume that the bolt does not experience air resistance. Calculate how long it takes for the bolt to hit the ?oor, how far it travels, and how fast it is going when it hits the ?oor of the elevator. Solve this problem in two reference frames, one ?xed with respect to the building and one ?xed with respect to the elevator. Only the fall time will be the same for the two reference frames. In this and subsequent problems you solve in dynamics, think about the advantages and disadvantages of various reference frames and clearly state what reference frame you ultimately choose for your solution.

Explanation / Answer

The bolt and the floor are moving up at 2 m/s (fixed WRT building) The floor is accelerating upward at 1 m/s^2. The bolt is accelerating downward at 9.8 m/s^2. We will solve for displacement and then equate the two displacements. The location of the floor right when the bolt falls out is 0

x bolt = 2.5 m + 2m/s*t - (1/2) 9.8m/s^2*t^2

x floor = 0 m + 2 m/s*t + (1/2)1 m/s^2*t^2

OK so when they hit x bolt = x floor, so we sub in the other sides of the equations since they are also equal.

2.5 +2t - 4.9 t^2 = 2t + .5 t^2

2.5 - 4.9 t^2 = .5 t^2

2.5 = 5.4 t^2

2.5/5.4 = t^2

0.68s = t

does this make sense? We start out going up at 2m/s whch adds to our time. Normally it takes 0.714s to fall 2.5m, but the floor is accelerating upward, so it should take less time.

So we are happy with

0.68s = time

Now let's find the distance by substituting 0.68s into the x bolt equation

x bolt = 6.12576 m,

but it started at 2.5 m, so it moved 3.6 m

Let's find the speed. Vf = Vo - gt = 2 m/s - 9.8m/s^2*0.68s = -4.7 m/

Now let's look in reference frame of the elevator.

Both are not moving. Then the bolt starts falling down toward the floor and the relative acceleration is 10.8 m/s^2

We want to know how long it takes to fall 2.5m (to the floor, which is position zero)

distance = starting distance + Vo*t - (1/2)at^2 = 0= 2.5m + 0 - (1/2)(10.8m/s^2)(t^2)

Like the question said, this is the same as in the other frame.

2.5/5.4 = t^2

0.68s = t

Distance: the floor "sees" the bolt move down 2.5 m. The floor doesn't see it is moving (even though it can feel itself accelerating)

Speed Here is the tricky part. Vf = Vo - at = 0m/s -10.8m/s^2*0.68s = 7.34 m/s

I like the elevator frame since it makes the starting speed zero.

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