PLEASE READ ALL THE WAY THROUGH!!! My question is at the end and is just a clari

ID: 1657785 • Letter: P

Question

PLEASE READ ALL THE WAY THROUGH!!!

My question is at the end and is just a clarification!

A 60.0 kg passenger is in an elevator that is accelerating Upward at 1.20 m/s2.

A) What is the apparent weight of the passenger?

Now for this part of the problem I got the answer 660N.

=660 N

THEN for part b it asks

b) What is the magnitude of the force of the passenger on the elevator floor?

Now, wouldnt part b just be the same as part a? Isn't apparent weight and magnitude of force the same thing?

If not, then please explain how I would find the answer for part b and why it is different.

Answer:-

Part-A] A) What is the apparent weight of the passenger?

As we have given the elevator is in UPWARD direction its acceleration is given a = 1.2 m/s2 Hence the net force

Fnet = m*a = 60 kg * 1.2 m/s2 = 72 N .

now determine the force due to gravity.

Fg = 60 *9.8 = 588 N

We know the formula for Net force

Fnet = normal force N -force due to gravity Fg

Hence the normal force N or apparent weight = 588+72 = 660 N.

Hence the Apparent weight of the passenger is 660 N.

b) What is the magnitude of the force of the passenger on the elevator floor?

if the direction of the acceleration is same to upward direction it will be the same as parta)

Because the net force will have same direction

N =588+72 = 660 N.

Hence the apparent weight on the elevator floor is 660 N.

(Note* if the direction of motion is downward in Part B then Net force will deduct from the force of gravity.

N = -72+588 = 516 N. )

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