Question

A Person whose weight is 5.20 x 10^2 N is being pulledup

Answer 1

Answer:

Explanation:

Given

Weight of Person $W=5.20\times 10^{2} N$

Cave is $h=35.1 m$ deep

Breaking stress $T=569 N$

Net Force on Person

$F_{net}=569-520=49 N$

$a_{net}=\frac{F_{net}}{\frac{W}{g}}$

$a_{net}=\frac{49}{\frac{520}{9.8}}$

$a_{net}=0.923 m/s^2$

The shortest time such that the person can be taken out of cave

$h=ut+\frac{1}{2}at^2$

where

h=distance moved

t=time

a=acceleration

$35.1=0+\frac{1}{2}(0.923)(t)^2$

$t^2=76.05$

$t=\sqrt{76.05}$

$t=8.72\ s$