Calculate the energy lost to friction

Thread starter okgo
Start date Mar 21, 2010
Tags

Energy Friction Lost

AI Thread Summary

The discussion focuses on calculating the energy lost to friction in a physics problem. The user initially seeks help with a specific homework question, relating the work done by friction to energy dissipation. They clarify that the work of friction can be expressed as umgcos(15)*d, where d is derived from the height and angle. After some calculations, the user resolves the issue independently, confirming their final expression for work done by friction. The thread highlights the process of solving physics problems involving friction and energy loss.

Mar 21, 2010

okgo

1. Homework Statement

http://i43.tinypic.com/i1bq1i.jpg
I need help with number 9

2. Homework Equations

Wnet=change in KE
Wnc=Change in Energy

The Attempt at a Solution

so I'm assuming the energy dissipated means the work done by friction?
so work of friction is umgcos(15)*d
d=cos(15)h
I got that work of friction is -9.8257
Help?

Physics news on Phys.org

Mar 21, 2010

okgo

nvm I figured it out.
.2*mgcos15*d
and the d=h/sin15
:approve:

Thread 'Errors and significant figures'

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...

View full post »

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...

View full post »

Thread 'Explain this asphalt sheen'

This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is The second part of the problem is exactly why you get this affect. And one more spoiler:

View full post »