A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10s. Then the coefficient of friction is

The correct option is 0.06.

To find the coefficient of friction, we can follow these steps:

Step 1: Identify the given information
- Mass of the marble block, $m=2\mathrm{kg}$
- Initial velocity of the block, $u=6m/s$
- Time taken to stop, $t=10s$
- Final velocity, $v=0m/s$ (since the block is stopped)
- Acceleration due to gravity, $g\approx 10m/s^2$

Step 2: Calculate the acceleration (deceleration) of the block
Using the first equation of motion:
$v=u+at$

Substitute the given values into the equation:
$0=6+a\left(10\right)$

$10a=-6$

$a=-0.6m/s^2$

The magnitude of the deceleration is:
$\left|a\right|=0.6m/s^2$

Step 3: Relate deceleration to the coefficient of friction
The retarding force acting on the block is the kinetic friction force ($f_k$):
$f_k=\mu N$
where $\mu$ is the coefficient of kinetic friction and $N$ is the normal force.

For a block on a horizontal surface, the normal force balances the weight of the block:
$N=mg$

Thus, the frictional force is:
$f_k=\mu mg$

According to Newton's second law, the force causing deceleration is:
$F=m\left|a\right|$

Equating the two forces:
$m\left|a\right|=\mu mg$

We can cancel mass ($m$) from both sides:
$\left|a\right|=\mu g$

Step 4: Solve for the coefficient of friction ($\mu$)
$\mu =\frac{\left|a\right|}{g}$

Substitute the values of $\left|a\right|$ and $g$:
$\mu =\frac{0.6}{10}=0.06$

Therefore, the coefficient of friction is 0.06.