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A stone of $1\, kg$ is thrown with a velocity of $20\, m\ s^{-1}$ across the frozen surface of a lake and comes to rest after travelling a distance of $50\, m$. What is the force of friction(in $N$) between the stone and the ice?
$-6$
$-4$
$-3$
$-2$
Solution
Initial velocity of the stone,$ u = 20\, m/s$
Final velocity of the stone, $v = 0\, m/s$
Distance covered by the stone, $s = 50\, m$
According to the third equation of motion:
$v^2 = u^2 + 2as$
Where,
Acceleration, $a$
$(0)^2 = (20)^2 + 2 \times a \times 50 $
$a = -4\, m/s^2$
The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone, $m = 1\, kg$
From Newton's second law of motion:
Force,
$F =$ Mass $\times $ Acceleration
$F= ma$
$F=1 \times(-4)=-4\, N$
Hence, the force of friction between the stone and the ice is $-4\, N.$
