(1 point) A hollow steel ball weighing 4 pounds is suspended from a spring. This
ID: 2891583 • Letter: #
Question
(1 point) A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring feet. The ball is started in motion from the equilibrium position with a downward velocity of 5 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 the positive direction is down.) Take as the gravitational acceleration 32 feet per second per second.Explanation / Answer
Solution:
Hooke's Law along with Newtons second Law of motion.
x" + (a/m)x' + (k/m)x = F(t)/m
x" + (160)x' + (96)x = 0
F = - kx
F = - 4
x = 1/4
mg = 4
m = 4/32
m = 1/8
- 4 = - k* (1/4)
k = 16
x" + (160)x' + (96)x = 0
y = C1e^(4*(640)(1/2) - 128)*t + C2e^(-4*((640)(1/2) - 128)*t)
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