A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms -1. The time the body takes to stop is

Question

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms -1. The time the body takes to stop is (A) 4 s (B) 6 s (C) 8 s (D) 12 s

Tardigrade Admin · Accepted Answer

a=(F/M)=(50 N /20 kg )=2.5 m / s 2 ⇒ t=(v-u/a) =((0-15/-2.5)) s =6 s

Q. A constant retarding force of $50 N$ is applied to a body of mass $20 k g$ moving initially with a speed of $15 m s^{- 1}$ . The time the body takes to stop is

4 s

6 s

8 s

12 s

$a = \frac{F}{M} = \frac{50 N}{20 k g} = 2.5 m / s^{2} \Rightarrow t = \frac{v - u}{a}$
$= (\frac{0 - 15}{- 2.5}) s = 6 s$

Q. A constant retarding force of 50N is applied to a body of mass 20kg moving initially with a speed of 15ms−1. The time the body takes to stop is

4 s

6 s

8 s

12 s

a=MF​=20kg50N​=2.5m/s2⇒t=av−u​ =(−2.50−15​)s=6s

Q. A constant retarding force of $50 N$ is applied to a body of mass $20 k g$ moving initially with a speed of $15 m s^{- 1}$ . The time the body takes to stop is

$a = \frac{F}{M} = \frac{50 N}{20 k g} = 2.5 m / s^{2} \Rightarrow t = \frac{v - u}{a}$
$= (\frac{0 - 15}{- 2.5}) s = 6 s$