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4-2.Friction
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A stone weighing $1$ kg and sliding on ice with a velocity of $2$ m/s is stopped by friction in $10$ sec. The force of friction (assuming it to be constant) will be ......... $N$
A$ - 20$
B$ - 0.2$
C$0.2$
D$20$
Solution
(b) $u = 2\;m/s,\;v = 0,\;t = 10\;\sec $
$\therefore $ $a = \frac{{v – u}}{t} = \frac{{0 – 2}}{{10}} = – \frac{2}{{10}} = – \frac{1}{5} = – 0.2\;m/{s^2}$
$\therefore $ Friction force $ = ma = 1 \times ( – 0.2) = – 0.2\;N$
$\therefore $ $a = \frac{{v – u}}{t} = \frac{{0 – 2}}{{10}} = – \frac{2}{{10}} = – \frac{1}{5} = – 0.2\;m/{s^2}$
$\therefore $ Friction force $ = ma = 1 \times ( – 0.2) = – 0.2\;N$
Standard 11
Physics
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