A 250 g block slides on a rough horizontal table. Determine the...

Given mass of the block, $m = 250 \text{ g} = 0.25 \text{ kg}$
Initial velocity, $v = 40 \text{ cm/s} = 0.4 \text{ m/s}$
Final velocity, $v_f = 0 \text{ m/s}$

By the work-energy theorem, the work done by the frictional force is equal to the change in kinetic energy of the block:
$W = \Delta K = K_f - K_i$
$W = 0 - \dfrac{1}{2} m v^2$
$W = - \dfrac{1}{2} \times 0.25 \times (0.4)^2 = -0.02 \text{ J}$

The frictional force acting on the block is given by $f = \mu m g$.
Taking $g = 9.8 \text{ m/s}^2$,
$f = 0.1 \times 0.25 \times 9.8 = 0.245 \text{ N}$

The work done by friction is also given by $W = -f d$, where $d$ is the distance moved before coming to rest.
$-0.02 = -0.245 \times d$
$d = \dfrac{0.02}{0.245} \text{ m} \approx 0.0816 \text{ m} = 8.16 \text{ cm}$

Rounding to one decimal place, the distance is $8.2 \text{ cm}$.