A 3 kg mass kept on a 30° slope is subjected to a variable force \(P = (t^2 + 3)\), where P is in ‘Newton’ and t is in ‘second’ find velocity of the block after 5 second using Impulse momentum principle. At t = 0, V = 0.[Figure: block on an incline r…

Given: m = 3 kg, W = 29.43 N, \(\mu = 0.2\), P = \(t^2 + 3\) N down the slope, u = 0. The block moves down the slope, so friction acts up the slope.

\(N = 29.43\cos30^\circ = 25.49\) N, \(F = 0.2 \times 25.49 = 5.10\) N; \(W\sin30^\circ = 14.715\) N

Net force down the slope: \(\sum F = t^2 + 3 + 14.715 - 5.10 = t^2 + 12.618\) N (positive at t = 0, so motion starts at once).

Impulse–momentum: \(mu + \displaystyle\int_0^5 \sum F\,dt = mv\)

\(\displaystyle\int_0^5 (t^2 + 12.618)\,dt = \dfrac{125}{3} + 63.09 = 41.67 + 63.09 = 104.75\) N·s

\(3v = 104.75 \Rightarrow v = 34.92\) m/s

Answer: v = 34.92 m/s down the slope after 5 s.