Coefficent of friction for a box moving at cosntant speed

[jwxie]

Homework Statement

A horizontal force fo 190N is needed to pull a 70kg box across a horizontal floor at a constant speed. What is the coefficent of friction between the box and the floor?

Homework Equations

Fnet = ma

The Attempt at a Solution

My setup is F - F_f = m*a

But I have trouble getting the acceleration...
F_f is supposed to be F_f = N * u

I am guessing F_f is knifectic fricition? Is it 190N?

I am confused. Please help!

 

[cepheid]

Homework Statement

A horizontal force fo 190N is needed to pull a 70kg box across a horizontal floor at a constant speed. What is the coefficent of friction between the box and the floor?

Homework Equations

Fnet = ma

The Attempt at a Solution

My setup is F - F_f = m*a

But I have trouble getting the acceleration...
F_f is supposed to be F_f = N * u

I am guessing F_f is knifectic fricition? Is it 190N?

I am confused. Please help!

Yes, F_f is the force of kinetic friction. Notice that the problem says that the box moves at a constant speed. So, what is the acceleration, then? Based on that, and Newton's Second Law (which you've written above), what must be true about how F_f compares to F (the applied force)?

 

[jwxie]

a = 0? so it means they are the same force??

So it means all we need is to use the Ff equation now.

 

[cepheid]

a = 0? so it means they are the same force??

So it means all we need is to use the Ff equation now.

Yes, a = 0, so the forces must be the same in magnitude. (They have opposite directions, so they cancel each other out).

Yeah, so since you know the value of F_f, and you know N, you can solve for μ

 

[Nickie]

The coefficient of friction is a measure of the resistance between two surfaces in contact. In this case, it is the resistance between the box and the floor.

To find the coefficient of friction, we can use the formula:

u = F_f / N

Where u is the coefficient of friction, F_f is the force of friction, and N is the normal force (the force exerted by the floor on the box).

In this problem, we are given the force of 190N needed to pull the box at a constant speed, and the mass of the box, 70kg. We can use the formula F = ma to find the acceleration of the box.

F = ma
190N = (70kg)a
a = 190N / 70kg
a = 2.71 m/s^2

Now, we can use this acceleration to find the force of friction, F_f.

F_f = ma
F_f = (70kg)(2.71 m/s^2)
F_f = 189.7N

Finally, we can plug in the values for F_f and N into the formula for the coefficient of friction:

u = F_f / N
u = 189.7N / (70kg)(9.8 m/s^2)
u = 0.27

Therefore, the coefficient of friction between the box and the floor is 0.27. This means that for every 1N of force applied to the box, there is 0.27N of resistance from the floor.