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A stone weighing 1 kg and sliding on ice with a velocity of $2 \mathrm{~m} / \mathrm{s}$ is stopped by friction in 10 s . The force of friction (assuming it to be constant) will be
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-0.2 N
$u=2 \mathrm{~m} / \mathrm{s}, \quad v=0, \quad f=0 \mathrm{~s}$
$a=\frac{v-u}{t}$
$=\frac{0-2}{10}=-\frac{2}{10}$
$=-\frac{1}{5}$
$=-0.2 \mathrm{~m} / \mathrm{s}^2$
$\therefore$ Friction force $=m a$
$=1 \times(-0.2)=-0.2 \mathrm{~N}$
$a=\frac{v-u}{t}$
$=\frac{0-2}{10}=-\frac{2}{10}$
$=-\frac{1}{5}$
$=-0.2 \mathrm{~m} / \mathrm{s}^2$
$\therefore$ Friction force $=m a$
$=1 \times(-0.2)=-0.2 \mathrm{~N}$
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