A person whose weight is 507 N is being pulled up vertically by a rope from the bottom of a cave that is 30.1 m deep. The maximum tension that the rope can withstand without breaking is 590 N. What is the shortest time, starting from rest, in which the person can be brought out of the cave?

Answer 1

W= mg=>
m=W/g = 507 /9.8 = 51.7 kg.

ma=T-mg
a= T/m - g =
=590/51.7 - 9.8 = 1.6 m/s²

h=at²/2
t= sqrt(2h/a) =
=sqrt(2•30.1/1.6) =6.13 s.

Answer 2

A person whose weight is 507 N is being pulled up vertically by a rope from the bottom of a cave that is 30.1 m deep. The maximum tension that the rope can withstand without breaking is 590 N. What is the shortest time, starting from rest, in which the person can be brought out of the cave?
force up = 590 constant so constant acceleration = a
force down = m g = 507 Newtons
net force up = 590 - 507 = 83 N
so
acceleration up = 83 / m = 83 / (507/g) = 83 / (507 / 9.81) = 83 / 51.7 = 1.61 meters/ second^2
distance = 1/2 a t^2 = 30.1 meters
(1/2) (1.61) t^2 = 30.1
t^2 = 37.4
t =6.11 seconds

Answer 3

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