 # Some more power stuff joe shmo Posers and Puzzles 04 Mar '10 04:16
1. 04 Mar '10 04:162 edits
I'm just stuck on a problem

the problem is asking for the power supplied by the motor at a certain time t.

The force the motor supplies is a function of time

F = F(t)

and

P = Fv

am I in the right ballpark for logic

F(t) = ma

a = F(t)/m = dv/dt

v= Int[(F(t)/m)dt] = v(t)

Then P(t) = v(t)* F(t)

??

The problem incorporates some pulleys and friction, but I didn't want to go into that if my logic is flawed from the get go.
2. 05 Mar '10 14:411 edit
You should offer up the problem.

Are you trying to determine power as a function of time given that you know the force as a function of time?
3. 05 Mar '10 20:57
Originally posted by AThousandYoung
You should offer up the problem.

Are you trying to determine power as a function of time given that you know the force as a function of time?
ok

a crate has a mass of 210 kg and rests on a(horizontal) surface for which the coefficients of static "s", and kinetic friction"k" are 0.4 & 0.3 respectivley

If the motor M supplies a cable force of F = (8t^2 + 20)N , where t is in seconds, determine the power output developed by the motor when t = 7 s.

This part you'll have to take my word

let +x axis be to the left

The FBD
3T to the left (because of mechanical advantage of pulleys) and Friction force to right.

sum Fy is not needed except for detrermination of Friction force which is just c(Fw)

c stands for coefficient, and Fw force weight.

there it is.
4. 06 Mar '10 10:41
Originally posted by joe shmo
am I in the right ballpark for logic

F(t) = ma

a = F(t)/m = dv/dt

v= Int[(F(t)/m)dt] = v(t)

Then P(t) = v(t)* F(t)
That looks the right approach to me.

You'll have to factor in the fact that the initial force isn't enough to overcome friction, so work out the time where the mass will start moving first. But then what you're doing here should work.