p08_016.pdf

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Chapter 8 - 8.16
16. We convert to SI units and choose upward as the + y direction. Also, the relaxed position of the top
end of the spring is the origin, so the initial compression of the spring (defining an equilibrium situation
between the spring force and the force of gravity) is y 0 =
−
0 . 100 m and the additional compression
brings it to the position y 1 =
−
0 . 400 m.
(a) When the stone is in the equilibrium ( a = 0) position, Newton’s second law becomes
F net = ma
F spring −
mg =0
−
k (
−
0 . 100)
−
(8 . 00)(9 . 8) = 0
where Hooke’s law (Eq. 7-21) has been used. This leads to a spring constant equal to k = 784 N/m.
(b) With the additional compression (and release) the acceleration is no longer zero, and the stone will
start moving upwards, turning some of its elastic potential energy (stored in the spring) into kinetic
energy. The amount of elastic potential energy at the moment of release is, using Eq. 8-11,
U = 1
2 ky 1 = 1
2 (784)(
−
0 . 400) 2 =62 . 7J .
(c) Its maximum height y 2 is beyond the point that the stone separates from the spring (entering free-
fall motion). As usual, it is characterized by having (momentarily) zero speed. If we choose the y 1
position as the reference position in computing the gravitational potential energy, then
K 1 + U 1 = K 2 + U 2
0+ 1
y 1 is the height above the release point. Thus, mgh (the gravitational potential
energy) is seen to be equal to the previous answer, 62 . 7 J, and we proceed with the solution in the
next part.
(d) We find h = ky 1 / 2 mg =0 . 800 m, or 80 . 0cm.
2 ky 1 =0+ mgh
where h = y 2 −

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