1. Use work-energy theorem

For the block moving from the top point to point $Q$,

$\Delta K = W_g + W_f$

where:

• $W_g$ = work done by gravity
• $W_f$ = work done by friction

Since the block starts from rest,

$K_i = 0, \qquad K_f = \frac12 mv^2$

So,

$\frac12 mv^2 = W_g - 150$

because 150 J is the work done against friction, hence friction does negative work:

$W_f = -150\text{ J}$

2. Find the vertical drop from the top to point $Q$

From the figure, point $Q$ is at the side of the circular arc, so the block descends through a vertical height equal to the radius:

$h = R = 40\text{ m}$

Therefore work done by gravity is

$W_g = mgh = (1)(10)(40) = 400\text{ J}$

3. Apply the energy equation

$\frac12 (1)v^2 = 400 - 150 = 250$

Thus,

$\frac12 v^2 = 250$

$v^2 = 500$

$v = 10\sqrt{5}\text{ m/s}$

4. Compare with options

The computed value is

$v = 10\sqrt{5}\text{ m/s} \approx 22.36\text{ m/s}$

This does not match any given option.

So the stored answer $\text{B }(10\text{ m/s})$ is inconsistent with the stated data.

If the intended vertical drop were $20\text{ m}$, then:

$mgh = 200\text{ J}$

$\frac12 v^2 = 200-150 = 50$

$v=10\text{ m/s}$

which would match option B. Hence either the figure/details are missing or the radius/position of $Q$ is not as inferred.

Therefore, based on the given text alone, the correct speed should be $10\sqrt{5}\text{ m/s}$, not one of the listed options.