Energy conservation from top to point $Q$

The block starts from rest, so initially $K_i = 0.$

Up to point $Q$, energy lost to friction is given as $W_f = 150\text{ J}.$

From the figure, the block moves from the top of the circular track to point $Q$ such that the vertical drop is the radius: $h = R = 40\text{ m}.$

Hence loss in gravitational potential energy is $mgh = (1)(10)(40) = 400\text{ J}.$

Therefore kinetic energy at $Q$ is $K_Q = 400 - 150 = 250\text{ J}.$

So, $\frac{1}{2}mv^2 = 250$ $\frac{1}{2}(1)v^2 = 250$ $v^2 = 500.$

Radial force balance at $Q$

At point $Q$, the radius is horizontal, so the inward radial direction is horizontal.

The weight $mg$ acts vertically downward, hence it has no radial component at $Q$.

Friction acts tangentially, so it also has no radial component.

Therefore, the only force providing centripetal force is the normal reaction $N$: $N = \frac{mv^2}{R}.$

Substituting the values, $N = \frac{(1)(500)}{40} = 12.5\text{ N}.$

Compare with options

The computed value is $N = 12.5\text{ N}.$

This does not match any option exactly. The closest option is $11.5\text{ N}$, but that is not equal to the calculated result.

Comparison with stored answer

Stored correct answer is A: $7.5\text{ N}$, which is inconsistent with the standard energy and centripetal-force analysis.

If the figure instead implies a different vertical drop, the result may change, but with the usual interpretation of point $Q$ as the side point of the circular arc, the correct value is $12.5\text{ N}$.

Since $12.5\text{ N}$ is not present in the options, the question/options likely contain an error.