Work Question involving Cheese, an elevator, and a cable

  • #1

Homework Statement


•••23 In Figure 7-34, 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?

http://www.lowellphysics.org/beta/Textbook%20Resources/Chapter%207.%20Kinetic%20Energy%20and%20Work/Problems/c07x7_11.xform_files/nw0316-n.gif


Homework Equations


Fn = Sum of forces down - Sum of forces up?
Wft = F*d*cos0


The Attempt at a Solution


Fn = 900g + .25g - Ft
3 = 900g + .25g
3 - (900 + .25)g = - Ft
Ft = 8828.5

WFt= 8828.5 (2.4) = 21188.4 J

Answer: 25900 J
 

Answers and Replies

  • #2
Doc Al
Mentor
45,204
1,541
Fn = Sum of forces down - Sum of forces up?
Only if the acceleration is zero.

The Attempt at a Solution


Fn = 900g + .25g - Ft
3 = 900g + .25g
3 - (900 + .25)g = - Ft
Ft = 8828.5

WFt= 8828.5 (2.4) = 21188.4 J
I don't understand what you're doing here. Hint: Find the acceleration of the elevator by analyzing the forces on the cheese.
 
  • #3
How does the acceleration help me though?

Find the acceleration of the cheese to use in the elevator system using

F = ma

And then solve for tension that way?
 
  • #4
Doc Al
Mentor
45,204
1,541
Exactly. The cheese and the elevator have the same acceleration.
 

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