Question

A stone of $1kg$ is thrown with a velocity of $20m{s}^{-1}$ across the frozen surface of a lake and comes to rest after travelling a distance of $50m$. What is the force of friction between the stone and the ice?

Answer 1

Step 1: Given

The initial velocity of the stone, u= $20m{s}^{-1}$

The final velocity of the stone, v= $0$

Distance covered by the stone, s= $50m$

Mass of stone, m = $1kg$

Step 2: Formula used

$v^2=u^2+2as$

$F=m\times a$, where $F$ is force.

Step 3: Calculation of acceleration and force of friction

Using,

$$\begin{gathered} v^2=u^2+2as \\ 0={20}^2+2\times a\times 50 \\ 100a=-400 \\ a=-\frac{400}{100} \\ a=-4m{s}^{-2} \end{gathered}$$

Hence, the acceleration of stone is $-4m{s}^{-2}$. Here, -ve sign shows retardation.

We know that

$F=m\times a$

Substituting the values in the above equation

$$\begin{gathered} F=m\times a \\ F=1\times (-4) \\ F=-4N \end{gathered}$$

Here, -ve sign shows force is in the opposite direction which is the frictional force.