91Ó°ÊÓ

The force applied to the block

$F=80\text{N}$

The distance traveled by the block while the force is being applied

$S_1=11\text{m}$

Time of application of force

$t_1=5\text{s}$

Time after which distance is to be calculated after the force is removed

$t_2=5\text{s}$

The initial velocity of the block is

u=0

The distance traveled s, time of travel t , initial velocity u and acceleration are related as

$S=ut+\frac{1}{2}a{t}^2.....\left(1\right)$

The second law of motion relates the force F , mass m and acceleration as

$F=ma.....\left(2\right)$

The final velocity, time of travel t, initial velocity u and acceleration are related as

$v=u+at.....\left(3\right)$

For uniform motion, the distance traveled is

$S=vt.....\left(4\right)$

Here, v is the velocity and t is the time.

Let the acceleration of the block be . From equation (1),

$\begin{array}{rcl}a& =& \frac{S_1-u{t}_1}{\frac{1}{2}{t}_1^2}\\ & =& \frac{11\text{m}-0}{\frac{1}{2}{\left(5\text{s}\right)}^2}\\ & =& 0.88{\text{m/s}}^{\text{2}}\\ & & \end{array}$

Thus from equation (2), the mass of the block is

$\begin{array}{rcl}m& =& \frac{F}{a}\\ & =& \frac{80\text{N}}{0.88{\text{m/s}}^{\text{2}}}\\ & =& 90.9\text{kg}\\ & & \end{array}$

The mass of the block is90.9kg.

From equation (3), the final velocity of the block is

$\begin{array}{rcl}v& =& u+a{t}_1\\ & =& 0+0.88{\text{m/s}}^{\text{2}}×5\text{s}\\ & =& 4.4\text{m/s}\\ & & \end{array}$

From equation (4), the distance traveled in time after the force is removed is

$\begin{array}{rcl}{S}_2& =& v{t}_2\\ & =& 4.4\text{m/s}×5\text{s}\\ & =& 22\text{m}\\ & & \end{array}$

Thus, the distance traveled is $22\text{m}$.