Time Taken for Block to Slide Down Moving Wedge

To find the time taken by block $X$ to slide down the wedge $Y$, we analyze the motion of the system using Newton's laws of motion.

1. Given Data

• Mass of wedge $Y$, $M = 10\text{ kg}$
• Mass of block $X$, $m = 2\text{ kg}$
• Pulling force, $f = 24\text{ N}$
• Angle of inclination, $\theta = 37^\circ \implies \sin 37^\circ = 0.6, \cos 37^\circ = 0.8$
• Distance to slide along the slope, $S = 8.8\text{ m}$
• Acceleration due to gravity, $g = 10\text{ m/s}^2$

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2. Horizontal Acceleration of the Wedge System

The total horizontal force $f = 24\text{ N}$ acts on the system consisting of the wedge and the block. The horizontal acceleration $a$ of the system is given by:

$a = \frac{f}{M + m}$

Substituting the given values:

$a = \frac{24}{10 + 2} = \frac{24}{12} = 2\text{ m/s}^2$

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3. Motion of Block $X$ Relative to Wedge $Y$

In the non-inertial reference frame of the accelerating wedge $Y$:

1. The component of acceleration due to gravity along the inclined plane (downwards) is: $a_{g, \parallel} = g \sin 37^\circ = 10 \times 0.6 = 6\text{ m/s}^2$

2. Since the wedge accelerates to the right with acceleration $a = 2\text{ m/s}^2$, block $X$ experiences a pseudo acceleration of $a = 2\text{ m/s}^2$ to the left. The component of this pseudo acceleration opposing the motion down the incline is: $a_{\text{pseudo}, \parallel} = a \cos 37^\circ = 2 \times 0.8 = 1.6\text{ m/s}^2$

Thus, the net relative acceleration $a_r$ of block $X$ sliding down the incline is:

$a_r = g \sin 37^\circ - a \cos 37^\circ$ $a_r = 6 - 1.6 = 4.4\text{ m/s}^2$

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4. Calculation of Time Taken

Using the kinematic equation for motion along the incline with initial relative velocity $u_r = 0$:

$S = \frac{1}{2} a_r t^2$

Substitute $S = 8.8\text{ m}$ and $a_r = 4.4\text{ m/s}^2$:

$8.8 = \frac{1}{2} \times 4.4 \times t^2$ $8.8 = 2.2 \times t^2$ $t^2 = \frac{8.8}{2.2} = 4$ $t = 2\text{ s}$

Thus, the time taken by block $X$ to slide down $8.8\text{ m}$ on the slope is $2\text{ s}$.