Energy conversation and friction

AI Thread Summary
The discussion revolves around a physics problem involving energy conservation and friction at a construction site. The initial calculations incorrectly placed dissipative work in the energy conservation equation, leading to an erroneous final speed for the bucket. The correct approach shows that the friction force on the box is 891.8 N, while the bag of gravel experiences no friction. After correcting the energy equation, the final speed of the bucket after descending 2.0 m is determined to be 2.99 m/s. The importance of accurately applying the energy conservation principles is emphasized throughout the discussion.
cdotter
Messages
305
Reaction score
0

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.
 
Last edited by a moderator:
Physics news on Phys.org
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top