# What is the impulse of friction?

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....

$$mv_{1} + F_{avg}\Delta{t} = mv_{}$$

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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 [Broken]

im trying to find the average Force at T = 8 seconds.....but its not working

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Doc Al
Mentor
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:
Doc Al
Mentor
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.