Question

I AM REALLY STRUGGLING ON THIS PROBLEM PLEASE HELP ME CORRECT AND NEAT WORK IS MUCH APPRECIATED THANKS

(1 point) A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring 13 feet. The ball is started in motion from the equilibrium position with a downward velocity of 6 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second). Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that this means that the positive direction for y is down.) y = Take as the gravitational acceleration 32 feet per second per second.

Answer 1

suspended from consider, a steel ball of weighting 128 pounds- mg of spring of spring const. K. elongation Al grave 128 17 feet accelaration, g232 ufft². 128 lb 4 lb mass of m = 32

The cond ?? ?? of eab? net Porce on here, acting forces - system = 0. ТКА. mg ma i) restoring force to = kal(upcourtd.) üy weight W= ( downward) sos (restoring force) - (weight)

KAL ? ma ма 128 pounds/fŁ 27 ka 128/17 k = 17 pounds Ift

F=-ka restoring force. resistive force Fr=-bà 220 Law of motion, Let, müa-ka-bå kim 6.2 b / 2m=8 → ä+028*+Wx 30 -ti) trial sol" areat » x=next ä=x²eat from (s), x² ext+280 ext two eat=0 y x² + 28xt we?=0 40 -8 I h. x= -8 I Joht L = √g tao? ret, CA, eht e e-ht). x=e8t t. x=A,(-8+h) ech-se +Az (-8-h) е - (8th)t at tao, a = 20 A, TAZ a) aado A, (7-8) + Az (-8-4) -8 oth (A1-A2) ci) from GY) + C7€) > t, -40/2 (1+ Boy **). Az = x/2 (+- -). Vol2 -8 h

here, nes resisting force, F :- bó? - 4 : 6= 4 pound -s/ft 2 :k/m 17 4 4,25 s 0.5 301 b. 2 m Leve Wo Yo Go, under damped oscillation, at theist) é 8t (A, where, js = vwo - peh 50, a = it A cos cetto) e 2. where A² - Ai + A2 bro + 2/2 -- ev) A sind ton & Acoso Now, 20 20 = 6 ft/s (4.25)? 2 h² 8?- we? (á) e- => -14.22 in = h); 4.22 51 2. 820 + V :-) 2 x2 + de A²+A from er of civ), 0:500 + 6 => A² 0²+ (. -)2 4.22 > Aadig from cu), tom & = brot Volk Xo put do 20 arget, tom & = of arget, tom & = o p = 1/2

x = e e-8€ A cos (at +¢). t/2 e- 1.19 cos e (4.22 € +T/2) et/2 totg sin (4.22€). -4 sign appears due to down ward motion. So, after t sec. position of balls tog sin (4:22t) ft y -ve at/2 Answer.