# Deceleration of an object due to friction

1. Nov 17, 2013

### keyofdoor

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Kinematic Equations

3. 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.

Last edited: Nov 17, 2013
2. Nov 17, 2013

### Lamebert

You found velocity, not a force.

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

Last edited: Nov 17, 2013
3. Nov 17, 2013

### keyofdoor

Can you detail on how I find the force needed?

4. Nov 17, 2013

### Lamebert

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.

5. Nov 17, 2013

### 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.)

6. Nov 17, 2013

### Lamebert

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.

7. Nov 17, 2013

### haruspex

You only need enough force to overcome static friction. Assuming it's a horizontal push, that's μsmg, 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).

8. Nov 17, 2013

### keyofdoor

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

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