# Homework Help: Help needed on Kinetic Energy and Work Problem

1. Oct 5, 2008

### ctpengage

1. The problem statement, all variables and given/known data

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?

2. The attempt at a solution

Solution Part 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,
F is the force from the cable, and 3.00 N N F = is the normal force on the cheese. On the
cheese alone, we have

FN - mg = ma
Therefore a = [3.00-(0.250)(9.80)] / 0.250 = 2.20 m/s2

Thus the force from the cable is F FTension = (m+M)(a+g) - FN = 1.08 x 104 and the work done by the cable on the cab is

W = FTensiond1 = (1.08 x 104)(2.40) = 2.59 x 104[/SUP

However I am stuck on Part B. The answer for Part B is 2.45 N. Can anyone please help me on Part B? Thanks guys.

2. Oct 5, 2008

3. Oct 5, 2008

### Staff: Mentor

The net force on the entire system (cab + cheese) would not include the normal force. The normal force is internal to the system and acts on both cab and cheese, but in opposite directions.

Good.

Redo this calculation using the correct net force.

Start by finding the cable tension. Then just work backwards, using the same approach as above.

4. Oct 6, 2008

### ctpengage

Why was my net force wrong?

5. Oct 6, 2008

### Staff: Mentor

See my last post: