Energy conversation and friction

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Homework Statement


At a construction site, a 65 kg bucket of concrete hangs from a light (but strong) cable that passes over a light friction-free pulley and is connected to an 80.0 kg box on a horizontal roof. The cable pulls horizontally on the box, and a 50.0 kg bag of gravel rests on top of the box. The coefficients of friction between the box and roof are shown.

A. Find the friction force on the bag of gravel and on the box.
B. Suddenly a worker picks up the bag of gravel. Use the energy conservation to find the speed of the bucket after it has descended 2.0 m from rest.

http://img237.imageshack.us/img237/8178/yf0729.jpg

Homework Equations



Ki + Ui + Wdissipative = Kf + Uf

The Attempt at a Solution



The frictional force on the gravel is 0 because it's not on the ground. For the box it's (130 kg)(9.8 m/s2)(0.7) = 891.8N.

Initial kinetic energy: 0
Initial gravitation potential energy: (65 kg)(9.8 m/s2)(2.0 m) = 1274 J
Dissipative work: (80.0 kg)(9.8 m/s2)(0.4)(2.0 m) = 627.7 J
Final kinetic energy: ?
Final gravitational potential energy: 0

Ki + Ui + Wdissipative = Kf + Uf
0 + 1274J + 627.7J = Kf + 0
Kf = 1901.7 J
1901 J = (145 kg*v2)/2 ==> v = 5.12 m/s

Where did I go wrong? By the kinematics equations, it should be 2.99 m/s.
 
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The two coefficents of friction are for standing and kinetic I hope.

only mistake i see is the "Ki + Ui + Wdissipative = Kf + Uf" If it is disspated it is on the right side.

Edited...
 
Lok said:
The two coefficents of friction are for standing and kinetic I hope.

only mistake i see is the "Ki + Ui + Wdissipative = Kf + Uf" If it is disspated it is on the right side.

Edited...

Yes, that's it. I guess I copied that equation down incorrect. Thank you very much. :smile:

Here are the corrected calculations for anyone else with this problem:

Ki + Ui = Kf + Uf + Wdissipative
0 + 1274J = Kf + 0 + 627.7J
Kf = 646.3J
646.3J = (145 kg*v2)/2 ==> v = 2.99 m/s