A wedge $Y$ with mass of 10 kg and all frictionless surfaces and the inclined surface making $37^{\circ}$ 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. \text { (Take } \tan \left(37^{\circ}\right)=3 / 4 \text { and } \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 \text { ) }