Time for a block to come to a stop on a horizontal surface

[Cocoleia]

Homework Statement

I have a block of mass m on a horizontal surface, which is covered in oil. The tell me the viscous resistance force is a function of the velocity, F(v)=-cv^1/2, where I am assuming c is a constant of some kind. I need to find the time that it will take for the block to stop. The initial velocity is V₀ at x=0

Homework Equations

Equations of motion
Newton's laws

The Attempt at a Solution

I thought that I could use ΣF=ma. Then in my equations of motion solve for a=(V-V₀)/t
Since the vertical forces cancel out, the equation would then be:
-cv^1/2=m((V-V₀)/t)
The problem is if I want the block to stop then V will be 0 and my equation doesn't work. Since the resistance is a function of V will I have to integrate it ?

 

[TSny]

Since the resistance is a function of V will I have to integrate it ?

Yes, integrate. The expression (V-V₀)/t represents the average acceleration over the time t. You need an expression for the instantaneous acceleration.

 

[Cocoleia]

Yes, integrate. The expression (V-V₀)/t represents the average acceleration over the time t. You need an expression for the instantaneous acceleration.

Do I also have to integrate F(V)=-cv^1/2 as well as the acceleration?

 

[TSny]

Do I also have to integrate F(V)=-cv^1/2 as well as the acceleration?

I don't follow. In your equation -cv^1/2=m((V-V0)/t) you need to replace the expression (V-V0)/t by an expression representing the instantaneous acceleration.

 

[drvrm]

I thought that I could use ΣF=ma. Then in my equations of motion solve for a=(V-V0)/t
Since the vertical forces cancel out, the equation would then be:
-cv1/2=m((V-V0)/t)
The problem is if I want the block to stop then V will be 0 and my equation doesn't work. Since the resistance is a function of V will I have to integrate it ?

as your force is velocity dependent write correct equation of motion
mass.rate of change of velocity = force and can then proceed further to solve it.

 

[Cocoleia]

I don't follow. In your equation -cv^1/2=m((V-V0)/t) you need to replace the expression (V-V0)/t by an expression representing the instantaneous acceleration.

What I'm asking is do I get the expression for the instantaneous acceleration by integrating the (V-V0)/t with respect to t ?

 

[TSny]

What I'm asking is do I get the expression for the instantaneous acceleration by integrating the (V-V0)/t with respect to t ?

No. Just use the definition of instantaneous acceleration.