Let $\mathbf{v}_1,$ $\mathbf{v}_2,$ $\dots,$ $\mathbf{v}_k$ be unit vectors such that
\[\mathbf{v}_1 + \mathbf{v}_2 + \dots + \mathbf{v}_k = \begin{pmatrix} 6 \\ -5 \end{pmatrix}.\]Find the smallest possible value of $k.$

Note: A unit vector is a vector of magnitude 1.
By the Triangle Inequality,
\[\|\mathbf{v}_1 + \mathbf{v}_2 + \dots + \mathbf{v}_k\| \le  \|\mathbf{v}_1\| + \|\mathbf{v}_2\| + \dots + \|\mathbf{v}_k\| = k.\]Then
\[k \ge \left\|\begin{pmatrix} 6 \\ -5 \end{pmatrix} \right\| = \sqrt{61} > \sqrt{49} = 7,\]so $k \ge 8.$

We can express $\begin{pmatrix} 6 \\ -5 \end{pmatrix}$ as the sum of 8 unit vectors, as shown below, so the smallest possible value of $k$ is $\boxed{8}.$

[asy]
usepackage("amsmath");

unitsize(1 cm);

int i;
pair A, B, C;

A = 6*(6,-5)/abs((6,-5));
C = (6,-5);
B = intersectionpoint(arc(A,1,-45,0),arc(C,1,90,180));

for (i = 0; i <= 5; ++i) {
  draw(i*(6,-5)/abs((6,-5))--(i + 1)*(6,-5)/abs((6,-5)),red,Arrow(6));
}
draw(A--B,red,Arrow(6));
draw(B--C,red,Arrow(6));

draw((-1,0)--(7,0));
draw((0,-6)--(0,1));

dot("$\begin{pmatrix} 6 \\ -5 \end{pmatrix}$", (6,-5), SE);
[/asy]