The answer is **no**. There can be instance in which *static* friction can do positive work. Consider two blocks, with one on top of the other. When the lower block is pulled a small force $F$, static friction stops the upper block from slipping, so static friction would apply a force parallel to $F$. The force of static friction and the displacement of the upper block are parallel, so the work done is positive.

In this problem, a coffee mug of mass $m = 0.12~mathrm{kg}$ is slid across a table over a distance of $d = 0.15~mathrm{N}$. The force exerted by the person is $F_{1} = 0.10~mathrm{N}$, and the coefficient of kinetic friction between the mug and the table is $mu_{k} = 0.05$. We calculate the work done on the mug. We use $g = 9.81~mathrm{m/s^{2}}$.

We first calculate the net force on the mug. The kinetic friction $f$ would point opposite the applied force. The normal force on the mug would be its weight $N = mg$. The net force would be

$$
begin{align*}
F &= F_{1} – f \
&= F_{1} – mu_{k}N \
F &= F_{1}
end{align*}
$$

The total work done on the mug is the dot product of the net force and the displacement.

$$
begin{align*}
W &= vec{mathbf{F}} cdot vec{mathbf{d}} \
&= Fd \
&= left( F_{1} – mu_{k}mg right)d \
&= left[ 0.10~mathrm{N} – left( 0.05 right)left( 0.12~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right)right] left( 0.15~mathrm{N} right)\
&= 6.171 times 10^{-3}~mathrm{J} \
W &= boxed{ 6.2 times 10^{-3}~mathrm{J} }
end{align*}
$$

$$
W = 6.2 times 10^{-3}~mathrm{J}
$$