Apparent weight word problem

Question

A 50kg student gets in a 1000kg elevator at rest. As the elevator begins to move, she has an apparent weight of 600N for the first 3s. How far has the elevator moved, and in which direction at the end of 3s?

Answer 1

Hi Essie!

 

Since you have posted a few of these, we can put them in context.  

 

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For a person standing in an elevator moving vertically, there are two principal forces at work: gravity, pulling the person down, and the normal force of the elevator floor, holding the person up.  With this realization, Newton's 2nd law always becomes (taking downward to be negative):

 

F_net = N - mg = ma

 

The person's "apparent weight," by the way, is just a code word meaning "the normal force on the person."  So, when you read "apparent weight," think "normal force."

 

If the apparent weight is greater than the person's "actual" weight, mg, the acceleration is positive (up).

If the apparent weight is less than the person's "actual weight," the acceleration is negative (down).

 

In other words, every time the person feels heavier, the acceleration must be up.  Every time the person feels lighter, the acceleration must be down.

 

[NOTE:  acceleration does not, by itself, indicate the direction of motion; under the right conditions -- which we can talk about in a bit -- the elevator could be moving up with either sign of acceleration, or down with either sign of acceleration]

 

Another fact to bring into this mix is that the condition of "speeding up," it turns out, occurs when velocity and acceleration have the same sign.  The condition of "slowing down" occurs when velocity and acceleration have opposite signs.

 

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So, for a problem like this, the first thing to look at is how the apparent weight compares to the actual weight.

 

Student's actual weight :  (50 kg)(9.8 m/s²) = 490 N

Student's apparent weight:  600 N

 

The student's apparent weight is greater than actual weight, so the acceleration must be UP (positive, as I defined it).  We are also told that the elevator "begins to move," which means it starts at rest and goes to some speed.  This is "speeding up."  Speeding up means the velocity must be in the same direction as the acceleration, which we know is up.  Therefore, the velocity must also be UP (positive).

 

This means the student has just begun being lifted.  This probably corresponds to your experience with elevators, where, upon starting upward from a lower floor, you felt heavier for a moment.

 

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We can use the knowledge of the apparent weight, and Newton's 2nd law, to determine the displacement.  This is because Newton's second law allows us to solve for acceleration, which we can then feed into kinematic expressions to solve for displacement.

 

Once again:

 

F_net = N - mg = ma

 

600 N - 490 N = (50 kg)a

110 = 50a

a = 110/50 = 2.2 m/s²

 

Then, in kinematic terms, we have:

 

to find:  (x-x_o)

know:  a (= 2.2 m/s²), v_o (= 0, starts at rest), t (=3.0 s)

 

Then choose the kinematic expression involving all these quantities, but not final velocity.

 

(x-x_o) = v_ot+(1/2)at²

(x-x_o) = (1/2)(2.2 m/s²)(3.0 s)² = 9.9 m (positive)

 

Technically, this is how far the student has moved, but we sure hope the student and elevator have moved the same distance!  Because of this, the mass of the elevator was not needed, unless we wanted to solve for something like the tension in the elevator cables.

 

Hope this helps!  If you have any other questions about this, just let me know.