What Do You Get When You Integrate Force?

[Alcubierre]

Hello,

I came across a problem the other day where the person integrated thrust force from 0 to y in respect to y. And that got me thinking: you integrate jerk to get acceleration and integrate acceleration to get velocity, so what do you get when you integrate a force, namely thrust force? Why would you even do such a thing?

 

[sankalpmittal]

Hello,

I came across a problem the other day where the person integrated thrust force from 0 to y in respect to y. And that got me thinking: you integrate jerk to get acceleration and integrate acceleration to get velocity, so what do you get when you integrate a force, namely thrust force? Why would you even do such a thing?

If we differentiate linear momentum with respect to time , we get force right ?

lim _Δt→0 Δp/Δt = dp/dt = F
F = dp/dt

So if we integrate Force with respect to time we get :

∫ F dt = ∫(dp/dt) dt
∫ F dt = p

which is linear momentum.

We use it as it has many uses.

You are given Force as a function of time :
F = t + 5t² + 6t³.
Now how will you obtain linear momentum of that body at t=5 ?

 

[mfb]

Hello,

I came across a problem the other day where the person integrated thrust force from 0 to y in respect to y. And that got me thinking: you integrate jerk to get acceleration and integrate acceleration to get velocity, so what do you get when you integrate a force, namely thrust force? Why would you even do such a thing?

What is y? A distance? In that case, you integrate force over a length, which gives energy.

 

[haruspex]

As you'll have seen from the other replies, it's not enough to say "integrate such-and-such". You have to specify the quantity you're integrating with respect to. For the purpose of understanding the nature of the answer, you can treat it like multiplication by that independent variable. If s, t are distance and time variables respectively, F.dt is force * time = momentum, F.ds is force * distance ('.' being the dot product of vectors) = energy, F\timesds is force-cross-product-distance = torque.