Q19E - University Physics with Modern Physics

Question

A 750.0-kg boulder is raised from a quarry 125 m deep by a long uniform chain having a mass of 575 kg . This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.50 times its weight without breaking. (a) What is the maximum acceleration the boulder can have and still get out of the quarry, and (b) how long does it take to be lifted out at maximum acceleration if it started from rest?

Step-by-Step Solution

Answer

(a) The maximum acceleration of the boulder is $0.832 \mathrm{m}/{\mathrm{s}}^2$ .

(b) The duration to come out from quarry is $17.3 \mathrm{s}$ .

Step 1: Given Data:

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 $T_m=2.5mg$  

Step 2: Acceleration:

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.

Step 3: Determine the acceleration of the system

(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 $M$,   $9.8 \mathrm{m}/{\mathrm{s}}^2$ for $g$ in the above equation.  

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

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

Step 4: Determine the duration to come out from quarry

(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.

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

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