A 4 kg block rests on a plane inclined at 20° to the horizontal. The coefficient of kinetic friction is 0.25. A variable force \(P=(2t+5)\) N, where t is in seconds, acts on the block parallel to the plane, pointing down the slope. If the block start…

Answer: \(v=13.2\) m/s down the plane.

\(W=4\times9.81=39.24\) N. Component of W down the plane \(=39.24\sin20^\circ=13.42\) N; normal reaction \(N=39.24\cos20^\circ=36.87\) N; kinetic friction \(F=0.25\times36.87=9.22\) N, acting up the plane (opposing motion down the plane).

Check that the block moves down from \(t=0\): at \(t=0\), force down the plane \(=5+13.42=18.42\) N \(>9.22\) N, so it starts moving down at once and friction stays kinetic.

Impulse–momentum along the plane (down the plane positive):

\[mv_1+\int_0^4\left(P+W\sin20^\circ-F\right)dt=mv_2\]

\[0+\int_0^4(2t+5)\,dt+(13.42-9.22)(4)=4v_2\]

\(\int_0^4(2t+5)\,dt=\left[t^2+5t\right]_0^4=16+20=36\) N s; \((4.20)(4)=16.81\) N s.

\[4v_2=36+16.81=52.81\Rightarrow v_2=13.2\ \text{m/s (down the plane)}\]