How Is Work Calculated in an Elevator Physics Problem?

[ctpengage]

Homework Statement

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?

2. The attempt at a solution

Solution Part 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 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

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

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

W = F_Tensiond₁ = (1.08 x 10⁴)(2.40) = 2.59 x 10<sup>4[/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.</sup>

 

[ctpengage]

Anyone please help?

 

[Doc Al]

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 is the force from the cable, and 3.00 N N F = is the normal force on the cheese.

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.

On the
cheese alone, we have

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

Good.

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

W = F_Tensiond₁ = (1.08 x 10⁴)(2.40) = 2.59 x 10⁴

Redo this calculation using the correct net force.

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.

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

 

[ctpengage]

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.

Why was my net force wrong?

 

[Doc Al]

Why was my net force wrong?

See my last post:

The net force on the entire system (cab + cheese) would not include the normal force.