A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. The force of friction between the stone and the ice is _______ N.

A stone of $1 \mathrm{kg}$ is thrown with a velocity of $20 \mathrm{m}/\mathrm{s}$ across the frozen surface of a lake and comes to rest after travelling a distance of $50\mathit{ m}$. The Force of friction between the stone and the ice is$-4 \mathrm{N}$.
Given:
Mass, $\mathrm{m} =\mathrm{ }1 \mathrm{kg}$
Initial velocity, $\mathrm{u} =\mathrm{ }20 \mathrm{m}/\mathrm{s}$
The final velocity, $\mathrm{v} =\mathrm{ }0 \mathrm{m}/\mathrm{s}$
Distance travelled, $\mathrm{s} =\mathrm{ }50 \mathrm{m}$
From the third equation of motion,
    ${\mathrm{v}}^{2\mathrm{ }}=\mathrm{ }{\mathrm{u}}^2\mathrm{ }+\mathrm{ }2\mathrm{as}$ $\Rightarrow \mathrm{a} =\mathrm{ }\frac{{\mathrm{v}}^{2\mathrm{ }}\mathrm{ }-\mathrm{ }{\mathrm{u}}^2}{2\mathrm{s}}$ $\Rightarrow$      $=\mathit{ }\frac{0 - {\left(20 \right)}^2}{2 \times 50 }$
$\Rightarrow$      $= \frac{-400}{100}$
      $\mathrm{a}= –4 \mathrm{m}/{\mathrm{s}}^2$
We know,
Force of friction between the stone and the ice is $=\mathrm{mass} \times \mathrm{acceleration}$.
                     $\mathrm{ }=\mathrm{ }1 \mathrm{kg} \times (-4) \mathrm{m}/\mathrm{s}$²        $=\mathrm{ }-4 \mathrm{N}$
Negative sign indicates that the direction of the force of friction is opposite to the direction of moving stone.