Big doubt about movement and friction

[MFDonadeli]

I have a doubt that I cannot resolve:

With a 10kg stop object on a surface I apply a 100 N force to move this object. The static friction coefficient is 0.5 and kinetic coefficient is 0.3. I would like to know the maximum speed of this object and the distanced covered till stop.

I know the only way the object moves is if the force applied is greater than static friction coefficient plus Normal force:

Femax = 0,5 x 10 x 9,8 = Femax = 49 N

When the object starts to move I have the applied Force (100 N) and the friction force:

100 - 0,3 x 10 x 9,8 = m x a -> 100 - 29,4 = m x a => 70,6 / 10 = a -> a = 7,06 m/s²

I found this acceleration, but how I know when the object will gain speed and when the object stops to accelerate and starts to stop. How the distance covered? How do I resolve this question?

This is not homework, it's a big doubt I have!

 

[chrisbaird]

If you are continually applying the force, as it sounds you are, then the acceleration will stay constant and positive. Acceleration is the change is speed with respect to time. The object will get faster and faster, gaining speed the whole time. It will not stop accelerating until you stop pushing. Then the decceleration is just that due to friction alone as there is no other force.

 

[MFDonadeli]

In this example the force is applied once. For example if I have a box on a table and I hit that box with my finger the box starts to move till stop.

In this case how do I know the distance covered (between start and end)?

 

[chrisbaird]

In this example the force is applied once...

Applying a force once does not make sense. Forces are applied over a period of time. Even if you flick a box, it looks instantaneous, but the force is being applied over a span of time. So we would have to know how long that time is in order to figure out to what speed the object has been accelerated to when the force goes away.

 

[MFDonadeli]

I think I understand now. Once I'm applying a force the object will be accelerating and the accelerator will decrease in the exact moment I stop to apply that force, and in this moment will be the maximum speed of the object. Is this correct?

 

[yster]

if the force is applied in the same manner as you hitting the box them the force becomes an impulse. Think about it- if you strike a ball with a bat the ball changes direction immediately and the rest of the bat's swing ends not include the ball.

If the force is applied gradually then the object will only accelerate to a certain velocity and maintain it because in reality friction is directly proportional to velocity.

 

[MFDonadeli]

I want to understand thinking about calculation:

In my example the acceleration a = 7.06 m/s². When I hit a ball or a box for a very short period (I don`t know, but for 0.2 seconds) then the final velocity is:

V = Vo + a x t => V = 0 + 7.06 x 0.2 -> V = 1.41 m/s (This is the maximum speed)

After this the force over this object is only the friction:

Fc = m x a => 0.3 x 10 x 9.8 = 10 x a => a = -2.94 m/s² (desaceleration)

And the total covered space is:

S= So + Vo x t + (at²)/2

In the time I'm applying the force:

S = 0 + 0 x 0.2 + (7.06 x 0.2) / 2 => S = 0.71 m

And after that thinking about desaceleration:

S = 0.71 + 1.41 + (2.94 x t) / 2 => S = 2.12 + (1.47 x t).

Obtening the t:

V = Vo + a x t => 0 = 1.41 + (-2.94 x t) => t = -1.41/2.94 => t = 0.48 s

The final distance will be:

S = 2.12 + (1.47 x 0.48) => S = 2.82 m

Is that correct?

 

[MFDonadeli]

Anyone can help me about the calculation? Please...