Finding Net Force and Position of Equilibrium for Moving Object

[~christina~]

Homework Statement

An object with mass m moves along the x axis. It's position as a function of time is given by

X(t)= At- Bt^2 + Ct^3

A,B,C are constants

a) find expression for the net force acting on the object as a function of time.

b) let A= 2m/s, B= 10m/s^2, and C= 1m/s^3

c) at what position is the net force 0?

d) Sketch graphs of the velocity and force vs time...

Homework Equations

F=ma

The equation given

The Attempt at a Solution

a) eqzn for the net force acting on the object as a function of time...

I have not a clue...is it I find the acceleration equation and then go and say m= constant so it doesn't matter?

x(t)=Ct^3 -Bt^2 + At

v(t)= x'(t)= 3Ct^2-2Bt +A

a(t)= v'(t)= 6Ct- 2B ==> is this it but how do I add m??

I assume since m is a constant so only thing changing is a, Am I correct?

b) when net force = 0
plugged in the numbers

x(t)= t^3 - 10t^2 + 2t

a(t)= 6Ct- 2B

a(t) 6t- 20
when is the force 0?? isn't it when accleration= 0?? and time is..

0= 6t-20
20/6= t

t= 3.33s

c) value of velocity when net force = 0

v(t)= x'(t)= 3Ct^2-2Bt +A

v(t)= 3t^2 - 20t + 2

t= 3.33 when net force = 0

plugging in..

v(3.33)= 3(3.33)^2- 20(3.33) + 2

v= -31 m/s

Is this alright?

 

[mjsd]

part a) F=ma so multiply m with the expression of a(t) to get force as a function of t
the rest seems ok

 

[~christina~]

what?
how would I multiply that into the a??
I just can't get my mind around to it...lets see

F= ma
a(t)= v'(t)= 6Ct- 2B

F(t)= m(6t- 20)

F(t)= 6tm- 20m ?

How can this part be wrong but the other parts okay?
wouldn't I get points taken off if the equation I was using didn't include the m for mass?
I would think so..

 

[mjsd]

you are confusing me with your question(s)...
You were asked to find F(t) given m and x(t) so as u have done, calculate a(t) then
F(t)=m a(t)
in part b) m cancels out anyway since it doesn't depend on t and so won't be affected by whether m is there or not, likewise for the subsequent parts.. except may be the graphing bit which you have not shown here anyway.

 

[Doc Al]

what?
how would I multiply that into the a??
I just can't get my mind around to it...lets see

F= ma
a(t)= v'(t)= 6Ct- 2B

F(t)= m(6t- 20)

F(t)= 6tm- 20m ?

That's all there is to it.

How can this part be wrong but the other parts okay?

Because it didn't ask for the force, but position and time where force was zero, which is where the acceleration is zero.

In part c, weren't you supposed to find position?

 

[~christina~]

I forgot to find position...

well since t= 3.33 s is when the force = 0

x(t)= t^3 - 10t^2 + 2t

so

x(3.33)= 3.33^3 - 10(3.33)^2 + 2(3.33)= -67.3m

but wait..why is distance negative...I calculated it twice..

how would the m cancel out in the
F(t)= 6tm - 20m??

when you plug in t would'nt you still have ..m?

Thank You

 

[stewartcs]

but wait..why is distance negative...I calculated it twice..

The distance is not negative, its position is negative.

The distance traveled will be positive even though the position is negative at the moment the acceleration is zero.

Don't confuse the total distance traveled (the displacement) with an objects position.

 

[Doc Al]

distance vs displacement

Don't confuse the total distance traveled (the displacement) with an objects position.

Just to nitpick on terminology a bit: I would say that the displacement (which is -67.3m, according to her calculations) does give you the final position (with respect to the starting point). Displacement is a vector describing change in position, whereas distance is a scalar representing the cumulative "ground" you've covered.

But your point remains. While "distance" traveled cannot be negative, the position (and displacement) certainly can.

 

[stewartcs]

I would have to say that you are indeed correct Doc Al, the displacement may certainly be negaitve. I should not have used displacement there, but rather distance.

Thanks for pointing that out.

 

[~christina~]

So the distance = 67.3m
but the displacement= -67.3m

Is this..what you people were saying?
If it is then I get it.

 

[mdk31]

So the distance = 67.3m
but the displacement= -67.3m

Is this..what you people were saying?
If it is then I get it.

Not necessarily. For example, if an object moves from position 0 m to position 300 m, and from position 300 m to a final position -290 m, the total displacement is -290 but the total distance traveled is certainly not +290. The total distance traveled would be 890

 

[Doc Al]

To add to what mdk31 already explained, you might want to read this: Distance and Displacement

 

[~christina~]

Not necessarily. For example, If an object moves from position 0 m to position 300 m, and from position 300 m to a final position -290 m, the total displacement is -290 but the total distance traveled is certainly not +290. The total distance traveled would be 890

so the displacement is just d= -67.3m but I don't know the distance? or do I?

To add to what mdk31 already explained, you might want to read this: Distance and Displacement

Thanks for the link

Hm..it explained vectors and scalars and displacement and distance however if it was just an equation like this problem was..I don't exactly have a picture of what is happening...

 

[Doc Al]

so the displacement is just d= -67.3m but I don't know the distance? or do I?

Since you have the position as a function of time, you can figure out the distance traveled between any two times. You'd have to examine the path it takes. (Make your own picture.) For example, at T = 0, the particle is at X = 0 and has a velocity of 2 m/s. Figure out how far it gets before it turns around, and just add up all the distance segments between each turn around point.

Not as simple as finding the final position, but you can do it.

 

[~christina~]

um..I don't think I need to find it for this problem so I'll work on my other ones first..however I ever need to find the distance traveled I know just what to do.

Thanks for explaining that