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

Question

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 (A) 2 s (B) 4 s (C) 6 s (D) 8 s

Tardigrade Admin · Accepted Answer

Here, F=-50 N (-ve sign for retardation) m=10 kg, u=10 m s-1, v=0 As F=m a ∴ a=(F/m) =(-50 N/10 kg) =-5 m s-2 using v=u+at ∴ t=(v-u/a) =(0-10 m s-1/-5 m s-2) =2 s

Q. A constant retarding force of $50 N$ is applied to a body of mass $10 k g$ moving initially with a speed of $10 m s^{- 1} .$ The body comes to rest after

$2 s$

$4 s$

$6 s$

$8 s$

Here, $F = - 50 N$ ( $- v e$ sign for retardation)
$m = 10 k g,$ $u = 10 m s^{- 1},$ $v = 0$
As $F = ma$ $∴$ $a = \frac{F}{m}$ $= \frac{- 50 N}{10 k g}$ $= - 5 m s^{- 2}$
using $v = u + a t$ $∴$ $t = \frac{v - u}{a}$ $= \frac{0 - 10 m s ^{- 1}}{- 5 m s ^{- 2}}$ $= 2 s$

Q. A constant retarding force of 50N is applied to a body of mass 10kg moving initially with a speed of 10ms−1. The body comes to rest after

2s

4s

6s

8s