Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / Section B / 6 / a / 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 slopeVertical balance gives $N=mg$, and horizontal motion gives $ma=-\mu mg$. ThusThe friction magnitude is $0.0300\mathrm{N}$.