v1kibc2

2023-02-04

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 has uniform strength, but it can only support a maximum tension of 2.50 times its weight at any point 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?

Keira Fitzpatrick

Mc=mass of the chain
Mb=mass of boulder
The weight of the chain is given by $Mc\cdot g=750\left[kg\right]\cdot 9.8\left[m/{s}^{2}\right]=7350\left[kN\right]$
Tension is pulling the boulder upward, while gravity is pulling it downward.
Sum of the forces=Mb*a
Mb*a=T-Mb*g
We know tension maxes out at 7350 [kN] so we will solve for the acceleration given that tension
a=(T-Mb*g)/Mb
$a=\left(7340\left[kN\right]-575\left[kg\right]\cdot 9.8\left[m/{s}^{2}\right]\right)/575\left[kg...m/{s}^{2}$
Now, using our position kinematic equation, we have
Yi=0
Yf=125 m
Vi=0
$a=2.96m/{s}^{2}$
t=?
$Yf=Yi+Vi\cdot t+\left(1/2\right)a\cdot {t}^{2}$
$125=0+0+\left(1/2\right)\left(2.96\right){t}^{2}$
$t=\sqrt{125/1.48}=9.19\left[s\right]$
A.) $2.96\left[m/{s}^{2}\right]$