Homework Help: HELP: Normal Force/Tension/Work Problem

1. Oct 3, 2008

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 FN = 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 FN?

The solution is:
(a) The net upward force is given by
Ftension + FN -(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,
Ftension is the force from the cable, and FN is the normal force on the cheese. On the cheese alone, we have
FN - mg = ma
Reaaranging and subbing in known values yields the acceleration as 2.20m/s2
Thus the force from the cable is
Ftension= (m+M) (a+g) − FN =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

FN = (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

Last edited: Oct 3, 2008
2. Oct 3, 2008

ctpengage

Re: Normal Force/Tension/Work Problem

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.

3. Oct 3, 2008