Given data

• Mass of block: $m = 10\,\text{kg}$
• Initial velocity: $u = 9.8\,\text{m s}^{-1}$
• Final velocity when it stops: $v = 0$
• Coefficient of friction: $\mu = 0.5$
• Acceleration due to gravity: $g = 9.8\,\text{m s}^{-2}$

Find the frictional force On a horizontal surface, normal reaction is $N = mg$ So friction is $f = \mu N = \mu mg$

Find the retardation produced by friction Using $f = ma$, $a = \frac{f}{m} = \frac{\mu mg}{m} = \mu g$ Since friction opposes motion, acceleration is negative: $a = -\mu g = -0.5 \times 9.8 = -4.9\,\text{m s}^{-2}$

Use the kinematic equation $v^2 = u^2 + 2as$ Substituting $v=0$, $0 = (9.8)^2 + 2(-4.9)s$ $0 = 96.04 - 9.8s$ $9.8s = 96.04$ $s = \frac{96.04}{9.8} = 9.8\,\text{m}$

Match with options $s = 9.8\,\text{m}$ So the correct option is B.

Comparison with stored answer Stored correct answer: B

My derived answer also gives B, so they agree.