Question

A body of mass 0.4 kg starting at origin at t = 0 with a speed of $10ms^{- 1}$in the positive x-axis direction is subjected to a constant F = 8 N towards negative x-axis. The position of the body after 25 s is :

• A. -6000 m
• B. -8000 m
• C. +4000 m
• D. +7000 m

Answer 1

Answer: (A)

-6000 m

Answer 2

Here

Mass of the particle$m = 0.4kg$

F = -8N

(minus sign for direction of force

$\therefore$Acceleration , $a = \frac{F}{m} = \frac{- 8N}{0.4kg} = - 20ms^{- 2}$

The position of the body at any time t is given by

$x = x_{0} + ut + \frac{1}{2}at^{2}$

The position of the body at t=0 is 0, therefore $x_{0} = 0$

$\therefore x = ut + \frac{1}{2}at^{2}$

Position of the body at $t = 25s$

Here $u = 10ms^{- 1},a = - 20ms^{- 2},t = 25$

$\therefore x = 10 \times 25 + \frac{1}{2}( - 20)(25)^{2}$

$= 250 - 6250 = - 6000m$