Kinetic friction on smooth then rough surface

[rasen58]

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 _____.

 

 

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.

 

[swansont]

I don't see a flaw in your reasoning.

[Klaynos]

Me neither. My back of the envelope gave the same numbers, 12N for the friction then 0.6 for the coefficient.

[rasen58]

Okay thanks, I guess the textbook gave the wrong answer.