Impulse. Change in Momentum vs Ft

[Ocata]

My book states that impulse = Ft = mv_f - mv_i

But it doesn't make sense to me because what if a Force of 5N is applied to an object for 10 seconds? Then I = Ft = 5N(10s) = 50 Ns.

So if I apply 5N to an object with a friction force of 5N for 10s so that the object is traveling at a constant velocity, say 3m/s, for the period of time, then nothing has changed in the calculation because Ft = 5N(10s).

However, if I make the same calculation using change of momentum, I get
Impulse = mvf - mvi = m(3m/s) - m(3m/s) = 0.

So why the different results if Ft = mv_f - mv_i ?

 

[Orodruin]

The impulse formula using the initial and final velocities are for the total impulse. If you have a friction force and travel at constant velocity, then it is not the only force acting on the object. The impulse from friction is Ft but there will be an equal but opposite impulse from the other force, making the total impulse zero. If you only have the friction force, you will not travel at constant velocity.

 

[Ocata]

If understand correctly, when the physics book (non calculus, applied physics) that I'm currently reading says Ft = mv_f - mv_i, what they specifically mean is F_net(Δt) = m(Δ v)?

So that when Force applied = Force of friction, Fnet = 0, so

F(t) = [(F_a - F_f) Δt = m(v_f - v_i)] = [(5N - 5N)10s = m(3m/s - 3m/s)] = [0 = 0]

That is, Ft = mvf - mvi = 0 when velocity remains constant due to Fnet = 0.

 

[Orodruin]

If understand correctly, when the physics book (non calculus, applied physics) that I'm currently reading says Ft = mv_f - mv_i, what they specifically mean is F_net(Δt) = m(Δ v)?

Correct.

 

[Ocata]

Thank you Orodruin