91Ó°ÊÓ

The mass of boulder is$M=750\mathrm{kg}$

The mass of uniform chain is $m=575\mathrm{kg}$

The depth of quarry is $d=125\mathrm{m}$

The maximum tension in the chain at a point is $T_m=2.5mg$

The maximum acceleration of the boulder can be found by equating the maximum tension at a point in chain by combined weight of the boulder and chain.

The second law of motion is given by,

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

Here $s$is the distance, $u$ is the initial speed, $t$ is the time and $a$ is the acceleration.

(a)

The maximum acceleration of the boulder is calculated as:

$T_m=\left(m+M\right)\left(g+a\right)$

Here, $a$ is the maximum acceleration of the boulder.

Substitute $2.5mg$for $T_m$, $575\mathrm{kg}$for $m$, $750\mathrm{kg}$for $M$, role="math" localid="1665055845384" $9.8\mathrm{m}/{\mathrm{s}}^2$for $g$in the above equation.

role="math" localid="1665056043928" $$\begin{gathered} 2.5mg=\left(\mathrm{m}+\mathrm{M}\right)\left(\mathrm{g}+\mathrm{a}\right) \\ 2.5\left(575\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)=\left(575\mathrm{kg}+750\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2+\mathrm{a}\right) \\ a=0.832\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the maximum acceleration of boulder is $0.832\mathrm{m}/{\mathrm{s}}^2$.

(b)

The duration to come out from quarry is calculated as:

$d=ut+\frac{1}{2}a{t}^2$

Here, $u$is the speed of boulder at rest and its value is zero.

Substitute 125 m for $d$, 0m/s for $u$and $0.832\mathrm{m}/{\mathrm{s}}^2$ for $a$in the above equation.

$$\begin{gathered} 125m=\left(0\right)t+\frac{1}{2}\left(0.832m/s^2\right)t^2 \\ t=17.3s \end{gathered}$$

Therefore, the duration to come out from quarry is 17.3 s .