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Q.
A constant retarding force of $50 \,N$ is applied to a body of mass $10\, kg$ moving initially with a speed of $10\, m\, s^{-1}.$ The body comes to rest after
Laws of Motion
Solution:
Here, $F=-50\, N$ ($-ve$ sign for retardation)
$m=10 \,kg,$ $u=10 m \, s^{-1},$ $v=0$
As $ F=m a$ $\quad\quad$ $\therefore \quad$ $a=\frac{F}{m}$ $=\frac{-50 N}{10 \, kg}$ $=-5 \, m \, s^{-2}$
using $v=u+at $ $\quad\quad$ $\therefore \quad$ $t=\frac{v-u}{a}$ $=\frac{0-10 \,m \,s^{-1}}{-5\, m \,s^{-2}}$ $=2\, s$