What happened to Force in this equation?

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SUMMARY

The discussion centers on calculating the deceleration of a 2000 kg car that skids to a stop due to kinetic friction. The coefficient of kinetic friction between rubber and pavement is established at 0.8. The normal force (N) is calculated as 19600 N, leading to a kinetic friction force (Fk) of 16000 N. The key takeaway is that the deceleration can be derived directly from the friction force using Newton's Second Law, where Fk equals the mass times acceleration (ma).

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Homework Statement



a 2000 kg car, after its brakes are locked up, skids and comes to a rest. The coefficient of kinetic friction between rubber and pavement is 0.8. Find deceleration of the car.




Homework Equations



M= 2000 kg
W= mg
V= 0




The Attempt at a Solution



N= mg = 2000 x 9.8 = 19600 N

Fk = N * Mk = 16000

F = ma

-Fk = 2000 a ...Where did F go? In most cases I thought F-Fk = ma. Here the F is missing. Is it because the car is skidding? I'm not sure and I was wondering if you guys could help me out.
 
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laxboi33 said:

The Attempt at a Solution



N= mg = 2000 x 9.8 = 19600 N

Fk = N * Mk = 16000

F = ma

-Fk = 2000 a ...Where did F go? In most cases I thought F-Fk = ma. Here the F is missing. Is it because the car is skidding? I'm not sure and I was wondering if you guys could help me out.

F-Fk=ma gives the resultant force.

Fk causes the car to decelerate so if we use ma=Fk, 'a' gives the deceleration of the car.
 
laxboi33 said:

Homework Statement



a 2000 kg car, after its brakes are locked up, skids and comes to a rest. The coefficient of kinetic friction between rubber and pavement is 0.8. Find deceleration of the car.




Homework Equations



M= 2000 kg
W= mg
V= 0




The Attempt at a Solution



N= mg = 2000 x 9.8 = 19600 N

Fk = N * Mk = 16000

F = ma

-Fk = 2000 a ...Where did F go? In most cases I thought F-Fk = ma. Here the F is missing. Is it because the car is skidding? I'm not sure and I was wondering if you guys could help me out.

Newton's Second Law:

[tex]\Sigma \vec F = m\vec a[/tex]

The sum of all forces is equal to the mass of the object times its acceleration.
For there to be a sum of forces, there doesn't have to be more than one force!

In this case, the only force acting on the car in the horizontal direction is the force of the kinetic friction, so:
[tex]\vec f_k = m\vec a[/tex]

The friction force in this case isn't a reaction force to an external force applied to the object, but it is present because there is relative motion between the car and the surface it is on.
 

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