What is the impulse of friction?

  • Thread starter Thread starter Jason03
  • Start date Start date
  • Tags Tags
    Friction Impulse
Click For Summary
SUMMARY

The discussion centers on calculating the impulse of friction in a momentum problem involving a block pushed by a force P over time T. The key equation used is mv1 + P - T(Δt) = mv2, where the impulse is defined as the area under the force-time curve. Participants emphasize the importance of correctly identifying forces, noting that P represents the applied force, while T is the kinetic friction force. The correct approach involves subtracting the impulse due to friction from the total impulse to find the net impulse acting on the block.

PREREQUISITES
  • Understanding of impulse and momentum concepts
  • Familiarity with force-time graphs
  • Knowledge of kinetic friction and its role in motion
  • Ability to calculate areas of geometric shapes (rectangles and triangles)
NEXT STEPS
  • Study the concept of impulse and its mathematical representation
  • Learn how to analyze force-time graphs for impulse calculations
  • Explore the role of kinetic friction in dynamics problems
  • Practice solving momentum problems involving multiple forces
USEFUL FOR

Students in physics, educators teaching mechanics, and anyone interested in understanding the dynamics of forces and motion, particularly in the context of impulse and friction.

Jason03
Messages
161
Reaction score
0
I was working on a basic momentum/impulse problem where a block is pushed by a force P over a time T that is plotted on a graph...Kinetic Friction is given but how would I incorporate that into the equation to find the velocity?

I know the impulse is the area under the curve...

mv_{1} + F_{avg}\Delta{t} = mv_{}
 
Physics news on Phys.org
What is the force of friction?
What is the impulse of friction?
How does that change your equation?
 
Actually I figured it out, pretty easy...I just over thought it as usual...
This equation allowed me to come up with the correct answer

mv_{1} + P-T(\Delta{t}) = mv_{2}
 
Im having trouble coming up with velocity for Time at 8 seconds...I came up with the Velocity at 5 seconds easily using the equation above...

heres the graph for the problem...http://img91.imageshack.us/img91/6935/graphmp3.jpg


im trying to find the average Force at T = 8 seconds...but its not working
 
Last edited by a moderator:
Show what you did. To find the impulse of the applied force, find the area under the curve. What about the impulse of the friction force?
 
I used this equation to solve for time at 5 seconds...and it worked...

mv_{1} + P-T(\Delta{t}) = mv_{2}


but I tried solving for time at 8 seconds... finding the area under the curve which is just a rectange and triangle...but the answer comes up incorrect

as far as area it should just be (L * W) + (.5*b*h)
 
Last edited:
Jason03 said:
I used this equation to solve for time at 5 seconds...and it worked...

mv_{1} + P-T(\Delta{t}) = mv_{2}
That equation is not quite right. That second term (P) is a force, not an impulse. (It should be the area under that curve.)


but I tried solving for time at 8 seconds... finding the area under the curve which is just a rectange and triangle...but the answer comes up incorrect

as far as area it should just be (L * W) + (.5*b*h)
Looks fine to me. Show me the details. What did you get for the impulse due to the force P? The impulse due to the friction force T?
 


Don't forget that Impulse is NET force * time. So if you have the graph of the force of the engine vs. time, you need to subtract out kinetic friction, which is considered constant over the time.

If the MR2 that kept up well with the viper was a Spyder, that is I think 137 hp and not 115 hp.
 

Similar threads

Replies
7
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
919
Replies
11
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
887
  • · Replies 8 ·
Replies
8
Views
2K