Calculating Impulse Based off a graph

  • Thread starter Thread starter doctorjuice
  • Start date Start date
  • Tags Tags
    Graph Impulse
AI Thread Summary
The discussion revolves around calculating impulse from a force graph for a 3.0 kg particle. The initial attempt involved summing the areas under the force curve but resulted in an incorrect value of 5000 N s. Participants noted that the error might stem from a misunderstanding of units, specifically confusing seconds with milliseconds. This realization led to a correction in the approach to calculating impulse. The conversation emphasizes the importance of unit consistency in physics calculations.
doctorjuice
Messages
7
Reaction score
0

Homework Statement



What is the impulse on a 3.0 particle that experiences the force shown in the figure (Figure 1) ?
Figure 1:
http://session.masteringphysics.com/problemAsset/1382988/2/09.EX04.jpg


Homework Equations


Impulse equations, force equations, how they relate.


The Attempt at a Solution


I added the area above the x-axis for the time interval 2 to 8 and subtracted the areas from 0 to 2 and 8 to 10. This answer came out to be 5000 N s which is incorrect. I'm confused because I thought the integral of force (the area bounded by the force graph) is equal to impulse.

Help would be greatly appreciated. :smile:
 
Physics news on Phys.org
Your result seems too large. s vs ms, perhaps?
 
lewando said:
Your result seems too large. s vs ms, perhaps?

Yes! I can't believe I didn't think of that. Thank you!
 
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