Elevator and spring scale question

In summary: So when the acceleration is in the opposite direction of the initial velocity, it's slowing down.In summary, the man in the elevator undergoes a series of accelerations and constant velocities. The spring scale registers a weight of 705.6N while the elevator is at rest, 813.6N during the first .800s of acceleration, and 7056N while traveling at a constant speed. To find the weight during the time interval when the elevator is slowing down, we need to find the man's acceleration during this time. This can be done by using Newton's second law, which states that force equals mass times acceleration. We know the man's mass (72.0 kg) and we can calculate the acceleration by using
  • #1
neshepard
67
0

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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  • #2
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).
 
  • #3
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?
 
  • #4
neshepard said:
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.
 
  • #5
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 without 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.
 
  • #6
neshepard said:
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.)
 
  • #7
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?
 
  • #8
neshepard said:
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).
 
  • #9
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.
 
  • #10
neshepard said:
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.
 

Related to Elevator and spring scale question

1. What is the purpose of an elevator and spring scale?

The purpose of an elevator and spring scale is to determine the weight of an object. The elevator scale measures the force exerted by the object due to gravity, while the spring scale measures the force required to stretch or compress the spring. Together, these two measurements can be used to calculate the weight of the object.

2. How do elevator and spring scales work?

Elevator and spring scales work by using the principle of Hooke's law, which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched or compressed. The elevator scale measures the distance the spring is stretched or compressed, while the spring scale measures the force required to stretch or compress the spring. By combining these measurements, the weight of the object can be determined.

3. Can elevator and spring scales be used to measure different units of weight?

Yes, elevator and spring scales can be calibrated to measure different units of weight, such as kilograms, pounds, or newtons. This is done by adjusting the scale's calibration to match the known weight of a standard object in the desired unit. For example, to measure in kilograms, the scale would be calibrated to match the weight of a 1 kilogram object.

4. Are elevator and spring scales accurate?

Yes, elevator and spring scales are generally considered to be accurate for measuring weight. However, their accuracy can be affected by factors such as the condition of the spring and the calibration of the scale. It is important to regularly check and calibrate these scales to ensure accurate measurements.

5. Can elevator and spring scales be used for heavy objects?

Yes, elevator and spring scales can be used for heavy objects. However, the maximum weight that can be accurately measured will depend on the specific scale being used. It is important to check the weight capacity of the scale before use to ensure it is suitable for the object being measured.

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