HELP: Normal Force/Tension/Work Problem

[ctpengage]

Hey there guys.

Our teacher gave us an example problem to which he provided the solution for us to study.

The question is:
A 0.250 kg block of cheese lies on the floor of a 900 kg elevator cab that is being pulled upward by a cable through distance d₁ = 2.40 m and then through distance d₂ = 10.5 m. (a) Through d₁, if the normal force on the block from the floor has constant magnitude F_N = 3.00 N, how much work is done on the cab by the force from the cable? (b) Through d₂, if the work done on the cab by the (constant) force from the cable is 92.61 kJ, what is the magnitude of F_N?

The solution is:
(a) The net upward force is given by
F_tension + F_N -(m + M)g = (m + M)a
where m = 0.250 kg is the mass of the cheese, M = 900 kg is the mass of the elevator cab,
F_tension is the force from the cable, and F_N is the normal force on the cheese. On the cheese alone, we have
F_N - mg = ma
Reaaranging and subbing in known values yields the acceleration as 2.20m/s²
Thus the force from the cable is
F_tension= (m+M) (a+g) − F_N =10800N and the work done is therefore
10800N x (2.4m) = 25900J

I understand part 1 but I don't understand the solution to part 2.

Solution Part 2:
If W = 92.61 kJ and d₂ =10.5 m , the magnitude of the normal force is

F_N = (m+M)g - W/d₂ = 2.45N

I don't understand the solution to part two at all. It seems that for part two, the acceleration of the cheese and elevator cab is zero. Why?

Can anyone help clarify issues with this problem.

Thanks

 

[ctpengage]

For guys who think I created some sort of fake scenario above to get you guys to help me cheat or something, the problem is no 25, chapter 7 from Fundamentals of Physics 8ed Extended and is a tutoring problem whose solution our prof. made available on WileyPLUS.

 

[ctpengage]

please guys help ere :D:D:D

 

[zackisdaboss]

We just got to this problem in my AP physics course. I thought the same thing as you when I arrived at the answer and checked it with the back of the book. The back of the book says that the normal force is equal to the weight of the cheese. The only way that is possible is if the accelerate or mass is zero. Since the mass is obviously not zero, it must be the acceleration. It then dawned on me that it must be because the acceleration of the cheese with the elevator cab as the frame of reference is zero. So then Fn= -mg = 2.45N. If you used no frame of reference the answer you should have arrived at is 4.902N.