Question

a force of hundred Newton acts on a body of mass 20 kg initially at rest for 5 seconds determine A final velocity attained by it be distance covered by it​

Answer 1

1. we know that F=ma from given data F=100N ,m=20kg ,t=5sec u=0(since intially at rest) v=? 100=20a; a= 5m/sec×sec We also know that v=u+at ( from Newton's laws of motion )

therefore ; final velocityv=0+5(5) =25m/sec.

Answer 2

(a) Final velocity, v = 25 m/s

(b) Distance, d = 62.5 m

Explanation:

Given that,

Force acting on the body, F = 100 N

mass of the body, m = 20 kg

Time, t = 5 s

(a) The force acting on a body is given by :

F = ma

$a=\dfrac{F}{m}$

$a=\dfrac{100}{20}$

$a=5\ m/s^2$

Initially the body is at rest, u = 0

Final velocity of a body is given by :

$v=u+at$

$v=at$

$v=5\times 5$

v = 25 m/s

(b) Let d is the distance covered by the body. It can be calculated as :

$v^2-u^2=2ad$

$v^2=2ad$

$d=\dfrac{v^2}{2a}$

$d=\dfrac{(25)^2}{2\times 5}$

d = 62.5 meters

Hence, this is the required solution.

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