Speed of Block Sliding Down Rough Circular Track

To find the speed of the block upon reaching point $Q$, we apply the Work-Energy Theorem to the block between the top of the track and point $Q$.

1. Geometry of the Track and Height Lost

From the given diagram:

• The track is a circular arc of radius $R = 40\text{ m}$.
• The center of the circular arc is at point $P(R, R)$.
• The block is released from rest at the top of the track, which is located at position $(0, R)$, so its initial height is $y_i = R$.
• The radius vector $PQ$ makes an angle of $30^\circ$ below the horizontal line passing through the top of the track and point $P$.

Therefore, the height of point $Q$ above the ground is: $y_Q = R - R \sin 30^\circ$

The vertical height lost by the block when sliding from the top of the track to point $Q$ is: $h = y_i - y_Q = R \sin 30^\circ = 40 \times \frac{1}{2} = 20\text{ m}$

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2. Work-Energy Theorem

According to the Work-Energy Theorem: $W_{\text{net}} = \Delta K$ $W_g + W_f = \frac{1}{2} m v^2 - \frac{1}{2} m v_i^2$

Where:

• $W_g = mgh$ is the work done by gravity.
• $W_f = -150\text{ J}$ is the work done by friction (since $150\text{ J}$ of work is done in overcoming friction).
• Initial velocity $v_i = 0$ (released from rest).
• $m = 1\text{ kg}$
• $g = 10\text{ m s}^{-2}$

Substituting the values into the equation: $(1)(10)(20) - 150 = \frac{1}{2} (1) v^2$

$200 - 150 = \frac{1}{2} v^2$

$50 = \frac{1}{2} v^2$

$v^2 = 100 \implies v = 10\text{ m s}^{-1}$

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Conclusion

The speed of the block upon reaching point $Q$ is $10\text{ m s}^{-1}$.

Correct Option: B