# Question #b38a1

Dec 28, 2016

$144 N$

#### Explanation:

Let magnitude of the force be $F$,
Using Newton's second law of motion we obtain constant acceleration $a$ as

$F = m a$

Inserting given value
$F = 16 a$
$\implies a = \frac{F}{16} m {s}^{-} 2$ ......(1)

Assuming that initially the body is at rest, final velocity $v$ of body after $3 s$ can be calculated with the help of kinematic expression

$v = u + a t$

Inserting given values and using equation (1) we get
$v = 0 + \frac{F}{16} \times 3$
$\implies v = \frac{3 F}{16} m {s}^{-} 1$ .....(2)

Once the force is removed the body moves at a constant velocity for next $3 s$. Distance $s$ covered in time is

$s = v t$

Inserting given values and using equation (2) we get
$81 = \frac{3 F}{16} \times 3$
$\implies \frac{F}{16} = 9$

Solving for force we get
$F = 9 \times 16 = 144 N$