The block's speed at the point where x=0.25A is v = 31.95 cm/s.
<h3>What is Spring constant?</h3>
The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies. The stiffer the spring is and the harder it is to stretch, the bigger the spring constant.
question is incomplete, this is the remaining statement
What is the amplitude of the subsequent oscillations? And What is the block's speed at the point where x=0.25A?
x = Asin(wt)
v = Aw coswt
at t = 0
w = sqrt(k/m)
v = Aw
A = v/w
A = 7.17 cm
part b )
E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2
mv^2 + k(1/4A)^2 = 1/2kA^2
mv^2 + kA^2/16 = kA^2
mv^2 = kA^2 - kA^2/16
mv^2 = 15kA^2/16
v^2 = 15/16 * (k/m) * A^2
v^2 = 15/16 *w^2A^2
v = sqrt(15/16) * wA
v = 31.95 cm/s
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<span>5 seconds expired during the deceleration.
Top rotated 45 radians during these 5 seconds.
First, calculate the chance in velocity, by subtracting the initial velocity from the final velocity. So
6 rad/s - 12 rad/s = -6 rad/s
So we lost a total of 6 rad/s. Divide that by the deceleration to give the number of seconds. So
-6 rad/s / -1.2 rad/s^2 = 5 s
So it takes 5 seconds for the deceleration to happen.
The equation that expresses the number of radians performed under constant deceleration with an initial velocity is
d = VT + 0.5 AT^2
where
d = distance
V = initial velocity
T = time
A = acceleration
Substituting the known values gives.
d = VT + 0.5 AT^2
d = 12 rad/s * 5s + 0.5 * -1.2 rad/s^2 (5s)^2
d = 60 rad -0.6 rad/s^2 *25s^2
d = 60 rad -15 rad
d = 45 rad
So the top rotated 45 radians while decelerating from 12 rad/s to 6 rad/s</span>
Answer:
Explanation:
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