A wedge \(Y\) with mass of \(10\) kg and all frictionless surfaces and the inclined surface making \(37°\) with horizontal. A block \(X\) with mass \(2\) kg is placed at the highest point of the wedge as shown in figure is at rest. At \(t = 0\) wedge \((Y)\) is pulled toward right with constant force \((f)\) of \(24\) N. Taking the block \(X\) at rest at \(t = 0\), the time taken by it to slide down \(8.8\) m on the slope, while \(Y\) is on the move, is _____ s.
(take \(\tan(37°) = 3/4\) and \(g = 10\) m/s\(^2\))

Horizontal acceleration of the wedge (assuming the block and wedge move together horizontally):
\(A = \dfrac{f}{M + m} = \dfrac{24}{10 + 2} = 2\text{ m/s}^2\)

Working in the non-inertial frame of the wedge, the block experiences a pseudo force \(mA\) directed opposite to the wedge's acceleration. The components along the incline are:
Component of gravity down the incline: \(g\sin 37^\circ\)
Component of pseudo force up the incline: \(A\cos 37^\circ\)

Net relative acceleration of the block down the incline:
\(a_{\text{rel}} = g\sin 37^\circ - A\cos 37^\circ\)

Using \(\sin 37^\circ = 0.6\) and \(\cos 37^\circ = 0.8\):
\(a_{\text{rel}} = 10(0.6) - 2(0.8) = 6 - 1.6 = 4.4\text{ m/s}^2\)

Time taken to slide a distance of \(8.8\text{ m}\) along the incline, starting from rest:
\(s = \dfrac{1}{2} a_{\text{rel}}\, t^2\)

\(8.8 = \dfrac{1}{2}(4.4)\,t^2\)

\(8.8 = 2.2\,t^2\)

\(t^2 = 4 \Rightarrow t = 2\text{ s}\)

Hence, the required time is \(2\text{ s}\).