Find the acceleration in terms of friction

[ScullyX51]

Homework Statement

A block of mass m lies on a horizontal table. The coefficient of static friction between the block and the table is µ_s. The coefficient of kinetic friction is µ_k, with µ_k < µ_s.
Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration a of the block after it begins to move.

Homework Equations

f=ma

The Attempt at a Solution

The answer to this example is a=µ_sg-µ_kg.
I am very confused on problems in which both kinetic and static friction are given. Can someone explain to me what this statement means? If the object is moving, why are we subtracting kinetic from static, should the kinetic be greater at this point?

 

[PhanthomJay]

Homework Statement

A block of mass m lies on a horizontal table. The coefficient of static friction between the block and the table is µ_s. The coefficient of kinetic friction is µ_k, with µ_k < µ_s.
Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration a of the block after it begins to move.

Homework Equations

f=ma

The Attempt at a Solution

The answer to this example is a=µ_sg-µ_kg.
I am very confused on problems in which both kinetic and static friction are given. Can someone explain to me what this statement means? If the object is moving, why are we subtracting kinetic from static, should the kinetic be greater at this point?

I don't believe you have clearly understood the problem. It is given that a force is applied to the block that is ju-u-u-st enough to get it to move. I mean like if it was just an iota less, it wouldn't move. What is the magnitude of that force? Once you establish it, that same force continues to be applied to the object. What is now the NET force acting on the moving object ? (remember that the block is moving, so kinetic friction is working agin' ya). Once you identify the NET force, solve for the acceleration using Newton 2.

 

[thomate1]

The statement is asking for the acceleration of the block after it begins to move. In this case, the block is initially at rest and the force of static friction is holding it in place. When you push on the block with enough force to overcome the force of static friction, it will begin to move. However, there will still be a force of kinetic friction acting on the block, which will cause it to decelerate.

In order to find the acceleration, we need to use the equation f=ma, where f is the net force acting on the block. In this case, the net force is the force you are applying minus the force of kinetic friction. This is because the force you are applying is greater than the force of kinetic friction, so there is a net force in the direction of motion.

The force of kinetic friction is equal to µkmg, where µk is the coefficient of kinetic friction and mg is the weight of the block. Therefore, the net force is (F-µkmg), where F is the force you are applying.

Using f=ma, we get (F-µkmg)=ma, and solving for a gives us a=(F-µkmg)/m. Since we know that the force you are applying is just enough to overcome the force of static friction, we can say that F=µsmg. Therefore, the final equation for acceleration is a=(µsmg-µkmg)/m, which can also be written as a=µsg-µkg.

In summary, the statement is asking for the acceleration of the block after it begins to move, which is affected by both the force you are applying and the force of kinetic friction. By subtracting the force of kinetic friction from the force you are applying, we get the net force acting on the block, which we can then use to find the acceleration.