Deceleration of an object due to friction

[keyofdoor]

Homework Statement

With how much force do I have to push a 1 kg object so that it goes 200 meters over a surface with a friction coefficient of 0.4?

Homework Equations

Kinematic Equations

The Attempt at a Solution

I listed the variables I knew off the bat
mass = 1 kg
distance = 200 meters
friction coefficient = 0.4
gravity = 9.82

I found a frictional force of 3.928 (therefore a deceleration of 3.928)

Using the equation :

$$V_f ^2 = V_i ^2 + 2ad$$
I put in :
$$0 = V_i ^2 + 2(-3.928)* 200$$
and by basic algebra:
$$V_i = 39.638...$$

And therefore you would need 39.638 Newtons of force to push a 1 kg object 200 meters with a friction coefficient of 0.4
I would like to know if this is correct and if not why am I wrong.

 

[Lamebert]

You found velocity, not a force.

Also, correct me if I'm wrong, but wouldn't this be work?

 

[keyofdoor]

You found velocity, not a force.

Can you detail on how I find the force needed?

 

[Lamebert]

Can you detail on how I find the force needed?

I'm not really sure, actually. I can find the work needed to move it 200 meters, which is just the frictional force times distance. If you want to push it over those 200 meters, at a constant speed, you can use 3.92N of force over the 200 meters, but I don't think a minimal amount of force applied to move it over a distance is a valid question.

I don't think kinematics equations are relevant if the only forces causing motion are friction and the push force.

 

[keyofdoor]

Sorry, I needed to clarify more,
If I only give it one push, without touching it ever again. Not a constant push.
(I'm very new to physics so I'm going to make a lot of mistakes.)

 

[Lamebert]

Sorry, I needed to clarify more,
If I only give it one push, without touching it ever again. Not a constant push.
(I'm very new to physics so I'm going to make a lot of mistakes.)

Even so, the force is still applied over some distance.

I'd just wait until someone who is more intelligent than I am to come in. It seems like a really simple problem, just as far as I know this would make more sense as a work problem.

 

[haruspex]

You only need enough force to overcome static friction. Assuming it's a horizontal push, that's μ_smg, your 3.928N. You appear to then divide that by the mass again to obtain 3.928ms^-2, which you termed a 'deceleration'. I suppose you could consider it a deceleration in this sense: if there were no friction and you applied a horizontal force F then the acceleration would be a = F/m; the friction reduces that by 3.928ms^-2.
But the answer to your question is just 3.928N. To cover 200m you just have to keep applying that force (or perhaps less if kinetic friction is lower).

 

[keyofdoor]

I'm pretty sure that's the answer, so thank you. Sorry for confusing you Lamebert.