Question

A body of mass 0.4 kg starting at origin at t =0 with a speed of 10 m/s in the positive x axis direction is subjected to a constant force F=8N towards negative x axis. Calculate the position of the particle for 25 seconds.

Answer 1

Hi....

mass = 0.4kg
force = 8N
acceleration = 8/0.4 = 20 m/s²

Assume direction from south to north as positive,
at the instant when the force acts,
u = 10 m/s
a = -20 m/s²
x₀ = 0
x = x₀ + ut + 0.5at²

At t = 25s,
x = 10×25 - 0.5×20×25² = -6000 m
So particle is at 6000m south of x=0 at t=25s...



Hope this helps u!!

Answer 2

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$\bold\red{hello...frd\:swigy\:here}$

$\bold\green{Your\:Answer:-}$

Mass of the body, m = 0.40 kg

Initial speed of the body, u = 10 m/s due north

Force acting on the body, F = –8.0 N

Acceleration produced in the body,

a = f/m = -8.0/0.40 = - 20 m/s^2

At t = –5 s

Acceleration, a' = 0 and u = 10 m/s

S = ut + 1/2 a' t^2 = 10× (-5)= -50 m

At t = 25 s

Acceleration, a'' = –20 m/s2 and u = 10 m/s

S' = ut' + 1/2 a'' t^2 = (10×25)+ 1/2 (-20)× (25)^2

= 250+ 6250= - 6000 m

I hope, this will help you

Thank you___❤

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