Why is the acceleration in the same direction as the net force and friction?

[undividable]

My question is simple, in the problem on the picture that i uploaded, why is -f_k =ma
I understand the friction is in the negative direction, só it is negative, but the netforce, and the aceleration, are also in the negative direction ,só why are they positive? Shouldt it be -fk=-ma ?

 

[robphy]

If you follow the rule that taking off the arrowhead gives the magnitude, you can write:
$-f_k=m a_x$ or $-f_k=-ma$.
That is, distinguish the "magnitude of the vector" $a=\left|\vec a\right|$ from its x-component $a_x$.

From the beginning...
In vector-form,
it's $\vec f_k \stackrel{this\ problem}{=}\vec F_{net} \stackrel{Newton II}{=} m\vec a$.
In component-form,
it's $(f_k)_x=F_{net,x}=ma_x$.

Based on the diagram, with the given friction force vector and the positive-x direction to the right,
this can be rewritten as
$-\left|\vec f_k\right|=ma_x$. ( So, since $m>0$, we expect $a_x<0$. )
The left-side is in terms of the magnitude of $\vec f_k$ and the right-side is in terms of the x-component of $\vec a$.

We could also write both sides in magnitude-form.
$-\left|\vec f_k\right|=m(-\left|\vec a\right|)$, or $-\left|\vec f_k\right|=-m\left|\vec a\right|$.

 

[Dale]

My question is simple, in the problem on the picture that i uploaded, why is -f_k =ma
I understand the friction is in the negative direction, só it is negative, but the netforce, and the aceleration, are also in the negative direction ,só why are they positive? Shouldt it be -fk=-ma ?

Use the same process I explained last time.