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Q5.10 A body of mass 0.40 kg moving initially with a constant speed of $10ms^{-1}$ to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be $t = 0$ , the position of the body at that time to be $x = 0$ , and predict its position at $t = -5 s, 25 s, 100 s$ .

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The acceleration of force is given by :

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

$a\ =\ \frac{-\ 8}{0.4}\ =-\ 20\ m/s^2$

At t = - 5 s :

There is no force acting so acceleration is zero and u = 10 m/s

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

or $=\ 10(-5)\ +\ \frac{1}{2}.0.t^2$

or $=\ - \ 50\ m$

At t = 25 s :

Acceleration is - 20 m/s² and u = 10 m/s

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

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

or $=\ -\ 6000\ m$

At t = 100 s

We have acceleration for the first 30 sec, and then it will move with constant speed.

So for 0 < t < 30 :

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

or $=\ 10(30)\ +\ \frac{1}{2}(-20)30^2$

or $=\ -\ 8700\ m$

Now for t > 30 s :

We need to calculate velocity at t = 30 sec which will be used as the initial velocity for 30 < t < 100.

$v\ =\ u\ +\ at$

or $=\ 10\ +\ (-20)30\ =\ -\ 590\ m/s$

Now $s\ =\ vt\ +\ \frac{1}{2}at^2$

or $=\ (-590)70\ +\ \frac{1}{2}.0.t^2$

or $=\ -\ 41300\ m$

Hence total displacement is : - 8700 + (- 41300 ) = - 50000 m.

Posted by

Devendra Khairwa

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