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Laws of Motion Part 2

Section: Physics

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Question : 4 of 52

Marks: +1, -0

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 \text{m/s}^2

[JEE Main 5 Apr 2026 Shift 1]

2
4
$\sqrt{2}$
$2\sqrt{2}$

Solution:

20 \sin 37^{\circ} - 2 a_{\omega} \cos 37^{\circ} = 2 a_{\text{rel}}

20 \cos 37^{\circ} - N + 2 a_{\omega} \sin 37^{\circ} = 0

24 - N \sin 37^{\circ} = 10 \times a_{\omega}

24 - (20 \cos 37^{\circ} + 2 a_{\omega} \sin 37^{\circ}) \times \sin 37^{\circ} = 10 a_{\omega}

24 - \left(16 + 2 a_{\omega} \times \frac{3}{5}\right) \times \frac{3}{5} = 10 a_{\omega}

24 - \frac{48}{5} - \frac{18}{25} a_{\omega} = 10 a_{\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}

20 \sin 37^{\circ} - 2 \times \frac{45}{29} \times \frac{4}{5} \times \cos 37^{\circ} = 2 a_{\text{rel}}

12 - \frac{72}{29} = 2 a_{\text{rel}}

\frac{348 - 72}{29} = 2 a_{\text{rel}}

\frac{276}{29 \times 2} = a_{\text{rel}} ; \frac{138}{29} = a_{\text{rel}}

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 2 \, \text{sec}

After rounding off

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