= Solution
After the push, the <forces> are weight $mg$ downward, <normal reaction> $N$ upward and sliding <friction> $F=\mu N$ opposite the <velocity>. There is no continuing applied push. The table has constant slope
$$
a=\frac{1.02-1.20}{1.50}=-0.120\,\mathrm{m\,s^{-2}}.
$$
Vertical balance gives $N=mg$, and horizontal motion gives $ma=-\mu mg$. Thus
$$
\boxed{\mu=\frac{0.120}{9.81}=0.0122.}
$$
The <friction> magnitude is $0.0300\,\mathrm N$.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1-forces.png]
{title=Free-body diagrams for horizontal sliding, the inclined parcel and the hanging weight}
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