A 400 lb weight is held 4 ft above a ledge. It is attached to a cable with a in
ID: 2996225 • Letter: A
Question
A 400 lb weight is held 4 ft above a ledge. It is attached to a cable with a in line shock absorber device of some kind. This cable is attached to the ground at the ledge. The weight free falls for 7.5 ft before the cable and shock absorber absorb any of the shock. The shock absorber absorbs 1000 lb in the first .45 seconds of the time in which it acts. After that, it absorbes 750 lb for the remainder of time of resonance in the fall until all the energy is absorbed. Calculate the distance in which it stretches past the original 7.5 ft before all the energy is absorbed into the shock absorber device. Show all work.
Explanation / Answer
W = 400 lb.
gain in K.E during free fall = loss in P.E = mg= Wh = 400 x 7.5 = 3000 lbft
i.e 1/2 m V^2 = 3000 lbft
1 lb = 32.2 lbm ft/s^2
and 1lb corresponds to 1 lbm where lb is pound force and lbm is pound mass
so m= 400 lbm
V = sqrt ( 3000 x 32.2 x 2 / 400 ) = 21.977 ft/s
when the shock absorber starts acting for the first time it absorbs 1000lb
that means a force of 1000lb acts on the weight in opposite direction to its motion i.e upwards.
so net force upwards is (1000_due to absorber ) - (400_due to gravity) = 600 lb
=> acc. a = - F/m = 600 x 32.2 lbm ft/s^2 / 400 lbm
= - 48.3 ft/s^2
=> velocity after 0.45 sec , V = Vo + at
= 21.977 - (48.3 x 0.45 )
= 0.242 ft /s
distance travelled = Vo t + 1/2 a t^2
= (21.977 x 0.45 ) + ( 0.5 x -48.3 x 0.45^2 )
= 4.999 ft
from then it absorbs onlyy 750 lb.
so net force upwards is only (750-400) = 350 lb
acc. a = - 350 x 32.2 / 400 = 28.175 ft/s^2
velocity at the starting of this 750 lb force is 0.242 ft/s
so when it stops its final velocity is 0.
V^2 - Vo^2 = 2as
=> 0 - (0.242^2) = 2 x -28.175 x S
=> S = 0.001 ft
so it stops at 5ft below the point where the free fall ends...
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.