Square $ ABCD $ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $ \triangle ABE $ is $40$. What is $ BE $? [asy]size(4cm);
pair A,B,C,D,E;
A=(0,0);
B=(0,50);
C=(50,50);
D=(50,0);
E = (30,50);
draw(A--B);
draw(B--E);
draw(E--C);
draw(C--D);
draw(D--A);
draw(A--E);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
label("$A$",A,SW);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,SE);
label("$E$",E,N);
[/asy]
The area of $\triangle ABC$ is $\tfrac{1}{2} \cdot AB \cdot BE.$ With $AB=10$, we get $\tfrac{1}{2} \cdot 10 \cdot BE = 40,$ or $5 \cdot BE = 40.$ Thus, $BE = \tfrac{40}{5} = \boxed{8}.$