# Elevator and spring scale question

## Homework Statement

A 72.0 kg man is standing on a spring scale in an elevator at rest. The elevator ascends, attaining it's max velocity of 1.20 m/s in .800s. It travels with this constant velocity for 5.00s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.5s and comes to rest. What does the spring scale register during the time interval it is slowing down?

## Homework Equations

There are other parts to this but I have those answers and I used T-mg=ma. I calculated the upward acceleration to be 1.5m/s^2. Other parts of the question asked what the scale read at rest, 705.6N, during the first .800s, 813.6N and at constant speed, 7056N.

## The Attempt at a Solution

Gotta say I have no clue how to solve for the last part, what does the spring scale register as it slows. I tried t=72.0kg(-1.5m/s^2)+72.00kg(9.8g) but that's wrong. Can somebody lead me in the right direction.

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Doc Al
Mentor
What's the man's acceleration as he slows? Then apply Newton's 2nd law. There are two forces acting on the man: his weight and the normal force of the scale (which is what the scale reads).

I understand that I need the acceleration as the man slows, but my question is how do I find that. Since technically he slows twice, once at the top and again at the bottom. I know his upward acceleration to a velocity, but not his downward acceleration, nor distance. That is why I put -1.5m/s^2 in the formula and that was wrong.

So how do I I go about determining his acceleration as he slows?

Doc Al
Mentor
Since technically he slows twice, once at the top and again at the bottom.
He only slows once. The first acceleration is him speeding up from 0 to 1.20 m/s.
I know his upward acceleration to a velocity, but not his downward acceleration, nor distance. That is why I put -1.5m/s^2 in the formula and that was wrong.
You know his initial and final speeds, and the time it took to reach that final speed. That's all you need to find the acceleration.

a=1.20-0/.800 = 1.5m/s^2

Maybe I'm missing something that is glaring, but after attaining max velocity of 1.2m/s there are no other speeds to work with except down.

so a=x-0/1.5 = unknown acceleration since I don't have the negative y acceleration.

I tried -1.5m/s^2 in the formula and that proved to be incorrect. Do I need to use the 5s somehow? If so, how? It seems to me that there is no deceleration implied with the statement of "travels with a constant speed for 5.00s." Does deceleration occur after 5.00s, during that time, is there a presumption I am to make somewhere that I am clueless about. I do know that apparently physics classes and textbooks are keen on "assume" this or that with out saying something. Assume your answer for projectile motion acceleration final is just before it hits the ground because when something hits the ground there is no more acceleration.

Doc Al
Mentor
a=1.20-0/.800 = 1.5m/s^2

Maybe I'm missing something that is glaring, but after attaining max velocity of 1.2m/s there are no other speeds to work with except down.
How long does it take for the elevator to slow down? That time is given. (You must be sliding right past it.)

Is the part I'm sliding past, the statement "The elevator then undergoes a uniform acceleration in the negative y direction for 1.5s and comes to rest" mean in the travel is down or the time to go from 1.2m/s to 0m/s?

Doc Al
Mentor
Is the part I'm sliding past, the statement "The elevator then undergoes a uniform acceleration in the negative y direction for 1.5s and comes to rest" mean in the travel is down or the time to go from 1.2m/s to 0m/s?
Yes, that's the time for the negative acceleration (the slowing down).

Awesome. Thanks.
How am I to know that a statement implying negative y direction is not really moving down, but slowing down? I know sometimes we put -9.8 for gravity, meaning down only because it comes out in the wash, but with this problem, it seems that I can take it to be up then down in travel.

Doc Al
Mentor
How am I to know that a statement implying negative y direction is not really moving down, but slowing down?
The acceleration is negative, not the velocity. In this problem the elevator only goes up!

When the acceleration and velocity are in the same direction (up in this case), the object is speeding up; when in opposite directions, it's slowing down.