Slope of an Energy with Friction Graph?

AI Thread Summary
The discussion centers on how the presence of friction affects the slope and y-intercept of a graph representing kinetic and potential energy versus work. Without friction, the slope of the graph is 1 and the y-intercept is 0. When friction is considered, if the work includes friction, the graph remains unchanged because the energy lost to friction is offset by the negative work done by friction. However, if friction is excluded, the slope would be less than 1 and the y-intercept would be below 0, reflecting the energy loss. Thus, the treatment of friction in the work calculation is crucial to understanding the graph's behavior.
Londoncalling
Messages
11
Reaction score
0

Homework Statement


An object is pulled at a constant F. KE0+PE0=0, so W=PE+KE. If there were no friction, the slope of a graph (KE+PE=Y-axis, W=X-axis) would be 1 and the y-intercept would be 0. What would the addition of friction do to the slope and y intercept (would the y intercept be more, less, or equal to 0 and the slope more, less, or equal to 1)

Homework Equations



KE0+PE0+W=PE+KE

The Attempt at a Solution


At first I thought that the slope would be less than 1 and the y intercept below 0, since my experimental results seemed to confirm this and since friction would be "taking" energy away. However, one of my classmates told me there should be no change. Can anyone explain this problem to me?
 
Physics news on Phys.org
It depends on how you define your work. If your work includes the work done by friction, then there would be essentially no change. If your work only includes work done by the original force, there should be a change.
 
Just so I can understand this--is this because friction would be taking away from both sides of the equation? Why "essentially" no change?
 
You have work done by both forces. Usually, we define the work done by your force to be positive. In that case, the work done by friction is negative (and is equal to the loss in KE and PE).

So, if you include friction, the graph wouldn't change. The PE+KE of the particle would go down because friction is converting that energy into heat, BUT, the negative work of friction would also bring your W down by the exact same amount. If you don't include friction, the graph will change. The work done by your force is the same, but the PE+KE of the particle will go down due to friction.
 
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