# Kinetic friction on smooth then rough surface

• rasen58
In summary, a 2.0 kg mass at rest on a horizontal surface experiences a constant force of 6 N for 30.0 m. For the first 15 m, the surface is frictionless, and for the second 15 m, there is friction. The coefficient of friction between the surface and the mass is 0.6, and the force of friction is 12 N to the left. This results in the mass coming to rest at the end of the 30 m. The correct answer for the coefficient of friction is 0.6, not 0.3.
rasen58

## Homework Statement

Given a 2.0 kg mass at rest on a horizontal surface at point zero. For 30.0 m, a constant horizontal force of 6 N is applied to the mass.

For the first 15 m, the surface is frictionless. For the second 15 m, there is friction between the surface and the mass.

The 6 N force continues but the mass slows to rest at the end of the 30 m. The coefficient of friction between the surface and the mass is _____.

## Homework Equations

F = ma
Work-kinetic energy theorem
F_friction = mu * m * g

## The Attempt at a Solution

To solve this, I found the velocity at 15 m by first using F=ma to find that the acceleration for the first 15 m is 3 m/s^2.

Then I used a kinematic equation to find that the velocity at 15 m is sqrt(90).
So then, for the second 15 m, I drew a force diagram and saw that for the mass to decelerate to 0 in the exact same distance as it took to accelerate, then the net force must be the same magnitude but in the reverse direction to slow it down.

So I thought that since it was previously just 6 N to the right, I thought the force of friction would have to be 12 N to the left so that the net force is 6 N to the left.
Force of friction = coeff_fric * m * g

So that means that 12 = coeff_fric * m * g

Solving for coeff_fric, I got 0.6.

But that is apparently wrong since it's supposed to be 0.3.
But the only way to get 0.3 is if Force of friction = 6 N to the left. But I don't see why it should be 6 N instead of 12 N to the left.

total force on the object in rough surface is, [(coffee-of friction)mg-F] coffee of friction = u
by work energy theorem, kinetic energy is lost in doing work against friction,
so F.x=1/2mv^2
F is = [(coffee-of friction)mg-F]
putting values we get
(20u-6)*15=1/2*90*2 ------>15 meters is length of rough surface

20u=12
u=.6
if u=.3 then both forces will balance each other so there will be no deceleration
and body will continue

@Sagar Singh coffee of friction = coefficient of friction?

I agree with 0.6

Sagar Singh
NascentOxygen said:
@Sagar Singh coffee of friction = coefficient of friction?

I agree with 0.6
haha typing mistake, coefficient of friction
i want to write, (coeff) but it auto corrected to (coffee) to coffee

Thank you everyone! Just wanted to confirm.

## 1. What is kinetic friction?

Kinetic friction is the force that resists the motion of an object as it slides over a surface. It is caused by the microscopic irregularities of the two surfaces in contact.

## 2. How does the smoothness of a surface affect kinetic friction?

A smoother surface will have less resistance to motion, resulting in lower kinetic friction. This is because there are fewer microscopic irregularities to create friction between the two surfaces.

## 3. How does kinetic friction change when moving from a smooth to a rough surface?

When moving from a smooth to a rough surface, the kinetic friction will increase. This is because the rough surface will have more microscopic irregularities, creating more points of contact and therefore more friction between the two surfaces.

## 4. Can kinetic friction be completely eliminated?

No, it is impossible to completely eliminate kinetic friction. Even on the smoothest surfaces, there will still be some microscopic irregularities that will create friction.

## 5. How is the coefficient of kinetic friction affected by a smooth then rough surface?

The coefficient of kinetic friction will increase when moving from a smooth to a rough surface. This is because the rough surface will have a higher coefficient of friction due to the increased number of points of contact and greater resistance to motion.

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