Displacement and Acceleration

[Raynor.Jim]

A 0.20 kg object moves along a striaght line. The net force acting on the object varies with the object's displacement. The object starts from rest at displacement X = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following.

A. The acceleration of the particle when its displacement x is 6 m

B. The time taken for the object to be displaced the first 12 m

C. The amount of work done by the net force in displacing the object the first 12 m

D. The speed of the object at displacement X = 12 m

E. The final speed of the object at displacement X = 20 m

F. The change in the moentum of the object as it is displaced from x = 12 m to x = 10 m

these equations are the ones I THINK are with the problem VF = Vo + at Vf2 = vo2 + 2ad (Final Velocity squared = Initial Velocity squared + 2 x acceleration x distance) D = Vot + 1/2 at2 ( Initial Velocity x Time + 1/2 x acceleration x time squared)



Im not really sure where to start with this as I have not really been explained very well what displacement and net force is.

Any help is appreciated.

 

[Redbelly98]

Welcome to PF.

Displacement is simply the position of the object. It starts "at 0", and ends up 20 meters away from the starting point.

But this problem, as written, appears to be unsolvable. There is not enough information given to answer these questions.

Finally, those equations you wrote will not apply, since they work when the acceleration has a constant value. However, the problem statement says the net force (and hence acceleration) is not constant, but varies with position.

Something is definitely missing here. Is the problem statement copied exactly as it was written?

 

[Raynor.Jim]

Yes that is everything.

 

[Redbelly98]

That is weird. In a problem like this, I would expect more information, such as the time (t=?) when the displacement has reached 20 m. Also, more information about the force, like whether it varies proportionally with displacement, or in some other fashion, or perhaps that it is in fact constant.

As it is, there really is no way of solving this.