PHY question.

[joonkimdds]

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hi guys, i was studying physics and i got stuck so i was wondering if someone can help me.


"A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.25 and the push imparts an initial speed of 3.9m/s?"

The solution says
Fnet = ma
0.25(9.8)M = MA <M cancels each other>
A = 2.45
and then they used delta X formula to get the answer.

my question is that if u guys imagine a box on the floor, there r normal force , gravity force, Friction force and then the force that pushed the box.
I know that normal force and gravity force cancel each other.
so when we set Fnet = MA, i think there should be Friction force and the push force on the left side, but solution only has Friction force on the left side.

0.25(9.8)M is the friction force that they wrote on the solution, but I say it should be "Force that pushes the box - Friction force = F net force."


where did i make mistake?

 

[DrBowtie]

Once the box is pushed the only force is the Ff acting to the left. When the box is pushed it has to overcome the static friction. Hence since we are talking about the kinetic friction the box is already moving.

 

[joonkimdds]

Once the box is pushed the only force is the Ff acting to the left. When the box is pushed it has to overcome the static friction. Hence since we are talking about the kinetic friction the box is already moving.

ah~~ thank u very much~!
but how can u know whether we r talking about the moment we push the box or right after we push the box?

 

[DrBowtie]

b/c we are talking about the time once the box has an initial velocity of 3.8 to stopping.

 

[Sicilian]

hi guys, i was studying physics and i got stuck so i was wondering if someone can help me.


"A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.25 and the push imparts an initial speed of 3.9m/s?"

The solution says
Fnet = ma
0.25(9.8)M = MA <M cancels each other>
A = 2.45
and then they used delta X formula to get the answer.

my question is that if u guys imagine a box on the floor, there r normal force , gravity force, Friction force and then the force that pushed the box.
I know that normal force and gravity force cancel each other.
so when we set Fnet = MA, i think there should be Friction force and the push force on the left side, but solution only has Friction force on the left side.

0.25(9.8)M is the friction force that they wrote on the solution, but I say it should be "Force that pushes the box - Friction force = F net force."


where did i make mistake?

You are trying to find the distance the box moved based on:

Fd = (1/2)(m)(v^2 final - v^2 initial)

Which you can rewrite as:

mad (yes mad 😡 ) = (1/2)(m)(v^2 final - v^2 initial)

You don't know the acceleration. But since you know the coefficient (co-efficient? 👍 ) of friction you can just use the equation for the frictional force to find it:

Fk = (uK)(Fn), where Fn is the normal force

Or ma = (uK)(mg), where mg is the normal force

So.... a = (uK)(g) = (0.25)(9.8) = 2.45

Now you can get mad again:

mad = (1/2)(m)(v^2 final - v^2 initial)

Since the m cancels, d = (0.5) (v^2 inital) / (a) = 3.1 m (assuming the final speed of the car is 0). You know the distance is positive, so don't kill yourself over signs (the authors of you're physics book don't do it either).

 

[Sicilian]

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