4, A force of 500 dyn is applied against the motion of a body of
mass 50 g moving at 1 m.s-1. When and where will the body
stop?
[Ans. 10 s; 5 m)
Answers
Answer:
let assume body is moving in positive direction of x - axis with initial speed u then the force F is applied against its motion( i.e negative direction of x-axis) applied such that it will stop after travelling distance s in time interval t
initial speed = u = 1m/s
then, final speed = v = 0m/s
net force = F = -500dyn = 500*10-^5 N (because 1dyne = 10^-50N )
Fnet = ma
Fnet = m(v-u)/t (if acceleration is constant )
-500*10^-5 = 50*10^-3(0–1)/t
t =- 50*10^-3/-500*10^-5
t = 10^-2-(-3)
t = 10 s
——————
s = ut + at^2/2
s = ut + (v-u)t^2/2t
s = 10 + (0–1)10/2
s = 10 - 5
s = 5m
Explanation:
hence we got (Ans. 10 s; 5 m) answer.
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Answer:
Explanation:
let assume body is moving in positive direction of x - axis with initial speed u then the force F is applied against its motion( i.e negative direction of x-axis) applied such that it will stop after traveling distance s in time interval t
initial speed = u = 1m/s
then, final speed = v = 0m/s
net force = F = -500dyn = 500*10-^5 N (because 1dyne = 10^-50N )
Fnet = ma
Fnet = m(v-u)/t (if acceleration is constant )
-500*10^-5 = 50*10^-3(0–1)/t
t =- 50*10^-3/-500*10^-5
t = 10^-2-(-3)
t = 10 s
____________
s = ut + at^2/2
s = ut + (v-u)t^2/2t
s = 10 + (0–1)10/2
s = 10 - 5
s = 5m
Hope it helps