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## 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