Question
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. $\left( \right.\ $ take $tan\left( 37^{\circ} \right) = 3/4$ and $\left. \ g = 10\text{ }m/s^{2} \right)$
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Solution
Sol. [Key Concept: From the wedge constraint find the block's acceleration relative to the ground along the incline; then $t = \sqrt{2s/a_{rel}}$.]
$${20sin37^{\circ} - 2a_{\omega}cos37^{\circ} = 2a_{\text{rel}\text{~}} }{20cos37^{\circ} - N + 2a_{\omega}sin37^{\circ} = 0 }{24 - Nsin37^{\circ} = 10 \times a_{\omega}}$$
$$\begin{matrix} & 24 - \left( 20cos37^{\circ} + 2a_{\omega}sin37^{\circ} \right) \times sin37^{\circ} = 10a_{\omega} \\ & 24 - \left( 16 + 2a_{\omega} \times \frac{3}{5} \right)\frac{3}{5} = 10a_{\omega} \\ & 24 - \frac{48}{5} - \frac{18}{25}a_{\omega} = 10a_{\omega} \\ & \frac{72}{5} = \left( 10 - \frac{18}{25} \right)a_{\omega} \\ & \frac{72}{5} = a_{\omega} = \frac{180}{116} = \frac{90}{58} \\ & a_{\omega} = \frac{45}{29} \\ & 20sin37^{\circ} - 2 \times \frac{45}{29} \times \frac{4}{5} \times cos37^{\circ} = 2a_{\text{rel}\text{~}} \\ & 12 - \frac{72}{29} = 2a_{\text{rel}\text{~}} \\ & \frac{348 - 72}{29} = 2a_{\text{rel}\text{~}} \\ & \frac{276}{29 \times 2} = a_{\text{rel}\text{~}};\frac{138}{29} = a_{\text{rel}\text{~}} \\ & t = \sqrt{\frac{2 \times 8.8 \times 29}{138}} \\ & t = \sqrt{\frac{2 \times 8.8 \times 29}{10 \times 138}} \\ & t = \sqrt{3.69} \\ & t \approx 2sec \end{matrix}$$
After rounding off
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