a) T - Mg = Ma
a is the upward acceleration, and you know what g is.
Solve for rope tension, T.
b) t = 2.0 seconds after the rope is cut, the helicopter will have risen an addional distance
D1 = V*t + (1/2)at^2
= 30*2 + (1/2)(5.2)*2^2 = 70.4 m
and the package will have risen an additional distance
D2 = 30*2 - (g/2)t^2 = 40.4 m
The difference is the separation.
Hi I really need this question answer as a check:
a helicopter holding a 70 kilogram package suspended from a rope 5.0 meters long accelerates upward at a rate of 5.2 m/s^2.
a. Determine tension on the rope.
b. When the upward velocity of the helicopter is 30 meters per second, the rope is cut and the helicopter continues to accelerate upwards at a rate of 5.2 m/s^2. Determine the distance between the helicopter and the package 2 seconds after the rope is cut.
3 answers
Thanx! that's what i got, other than the fact that we had a 5 foot difference with the rope already so the total should be 35 m after 2 secs.
bingo