Given data

• Mass of block: $m = 2\,\text{kg}$
• Initial velocity: $u = 6\,\text{m/s}$
• Final velocity: $v = 0$ (it stops)
• Time taken: $t = 10\,\text{s}$

Find retardation using first equation of motion

We use: $v = u + at$ Substituting the values: $0 = 6 + a(10)$ $10a = -6$ $a = -0.6\,\text{m/s}^2$

So, the magnitude of retardation is: $|a| = 0.6\,\text{m/s}^2$

Relate retardation to friction

On a horizontal surface, friction force is: $f = \mu N = \mu mg$

This friction produces the retardation: $f = ma$

Therefore, $\mu mg = m|a|$ $\mu g = |a|$ $\mu = \frac{|a|}{g}$

Substitute $g$

Taking $g \approx 10\,\text{m/s}^2$, $\mu = \frac{0.6}{10} = 0.06$

Match with options

$\mu = 0.06$ So the correct option is D.

Comparison with stored answer

Stored correct answer: D

Our derived answer is also D, so they agree.