Normal Reaction Acting on Block at Point Q

To find the magnitude of the normal force exerted on the block at point $Q$, we analyze the geometry of the track and apply the principles of circular motion and the Work-Energy Theorem.

Step 1: Geometry of the Track and Position of Point $Q$

The circular track has a radius $R = 40\text{ m}$ with its center located at $P(R, R)$.

• At the top of the track, the block starts from rest at height $y_i = R$.

• Point $Q$ lies on the circular arc such that the radius vector $PQ$ makes an angle of $30^\circ$ below the horizontal line $PR$.

• The vertical coordinate of point $Q$ is given by: $y_Q = R - R \sin 30^\circ$

• Thus, the vertical distance fallen by the block from the top to point $Q$ is: $h = y_i - y_Q = R \sin 30^\circ = 40 \times \sin 30^\circ = 20\text{ m}$

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Step 2: Determination of Speed at Point $Q$

By the Work-Energy Theorem, the total work done on the block as it moves from the top to point $Q$ equals the change in its kinetic energy:

$W_{\text{gravity}} + W_{\text{friction}} = \frac{1}{2} m v^2 - 0$

Given:

• Mass $m = 1\text{ kg}$
• $g = 10\text{ m s}^{-2}$
• $W_{\text{gravity}} = mgh = (1)(10)(20) = 200\text{ J}$
• Work done against friction is $150\text{ J}$, so $W_{\text{friction}} = -150\text{ J}$

Substituting these values into the equation: $200 - 150 = \frac{1}{2}(1)v^2$ $50 = \frac{1}{2}v^2 \implies v^2 = 100\text{ m}^2\text{s}^{-2}$

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Step 3: Calculation of Normal Force at Point $Q$

At point $Q$, the forces acting on the block along the radial direction (towards the center $P$) are:

1. The normal force $N$ directed radially inwards towards $P$.
2. The radial component of gravity pointing radially outwards away from $P$, given by $mg \sin 30^\circ$.

The net force directed towards the center provides the required centripetal force for circular motion:

$N - mg \sin 30^\circ = \frac{m v^2}{R}$

Rearranging to solve for $N$:

$N = mg \sin 30^\circ + \frac{m v^2}{R}$

Substituting the known values:

$N = (1)(10)\left(\frac{1}{2}\right) + \frac{(1)(100)}{40}$ $N = 5 + 2.5 = 7.5\text{ N}$

Thus, the magnitude of the normal force exerted on the block at point $Q$ is $7.5\text{ N}$.