What is the x-component of the force on the particle at x=5, 15, 25, and 35 cm
i thought it was just the slope?!? but im wrong ,some help?
x=15 i know F=0
Attachments
graph.jpg
10.5 KB
· Views: 993
Answers and Replies
#2
Brad Barker
429
0
psingh said:
What is the x-component of the force on the particle at x=5, 15, 25, and 35 cm
i thought it was just the slope?!? but im wrong ,some help?
x=15 i know F=0
so close!
there's a formula that should be in your text:
|F| = - dU/dx. (direction is along the x-axis.)
(if, like in this problem, we are only concerned with one dimension.)
in three dimensions, F = - grad U
#3
psingh
18
0
oye you're right neg. slope =] thankss
#4
Dr.Brain
538
2
In differential form , the relation is fiven by
[itex]F = \frac{-dU}{dR}[/itex]
Just find the slope of the tangents at each of the given x-coordinates . The answer will be positive or negative depending on whether Potential is decreasing or increasing at that point.