Answer
$0.14\,m/s^{2}$
Work Step by Step
It is the x component of the applied force that provides the acceleration. Some of this force is utilized to overcome friction. Using Newton's second law, we can write
$F_{x}-f_{k}=ma$
That is, $15\,N\times\cos40^{\circ}-11\,N=3.5\,kg\times a$ (In the previous section, we found $f_{k}$ to be equal to 11 N)
$\implies a=\frac{0.50\,N}{3.5\,kg}=0.14\,m/s^{2}$
