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**Unformatted text preview: **Chapter 8 1. The potential energy stored by the spring is given by U = §loc2 , where k is the spring constant and x is the displacement of the end of the spring from its position when the
spring is in equilibrium. Thus k — 393233—2939). 103N/m.
x (0.07m) . 3. (a) Noting that the vertical displacement is 10.0 m - 1.50 m = 8.50 m downward (same direction as F; ), Eq. 7-12 yields
Wg = mgd cos ¢ = (2.00 kg)(9.80 m/s2)(8.50 m) cos 0° = 167 J, (b) One approach (which is fairly trivial) is to use Eq. 8-1, but we feel it is instructive to
instead calculate this as AU where 'U = mgy (with upward understood to be the +y
direction). The result is AU = mg(yf — y;) = (2.00 kg)(9.80 m/s2)(1.50 m—10.0 m) = —167 J. (c) In part (b) we used the fact that U,- = mgyi =196 J.
((1) In part (b), we also used the fact Uf= mgyf= 29 J. (c) The computation of Wg does not use the new information (that U = 100 J at the
ground), so we again obtain Wg = 167 J. (t) As a result of Eq. 8-1, we must again ﬁnd AU = —Wg = —167 J. (g) With this new information (that U0 = 100 J where y = 0) we have
U,- = mgyi + U0 = 296,131 (h) With this new information (that U0 = 100 J where y= 0) we have
Uf= mgyf+ U0 = 129 J. We can check part (f) by subtracting the new U,- from this result. 15. We neglect any work done by friction. We work with SI units, so the speed is
converted: v = 130(1000/3600) = 36.1 m/s. (a) We use Eq. 8—17: Kf + Uf = K,- + U,- with U,- = 0, U,» = mgh and Kf = 0. Since
K. = émv2 , where v is the initial speed of the truck, we obtain 1 2 2
—1—mv2=mgh :> h=L=—(—3—6—1I—n/——SZ—=66.5m.
2 2g 2(9.8 m/s ) IfL is the length of the ramp, then L sin 15° = 66.5 In so that L = (66.5 m)/sin 15° = 257
m. Therefore, the ramp must be about 2.6 x102 m long if ﬁ'iction is negligible. (b) The answers do not depend on the \mass of the truck. They remain the same if the mass is reduced. /
(c) If the speed is decreased, 11 and L both decrease (note that h is proportional to the
square of the speed and that L is proportional to h). 56. Energy conservation, as expressed by Eq. 8-33 (with W: 0) leads to AEm =Ki—Kf+Ui—Uf :> fkd=0—O+%kx2—O 1 ‘ '-
:> ,ukmgd=§(200N/m)(0.15m)2 => ,uk(2.0kg)(9.8m/s2)(0.75m)=2.25 J \
k\§ which yields ,uk = 0.15 as the coefﬁcient of kinetic friction. 60. We look for the distance along the incline d, which is related to the height ascended
by Ah = d sin 6. By a force analysis of the style done in Chapter 6, we ﬁnd the normal
force has magnitude F N = mg cos6l, which means ﬁ = ,uk mg 0056. Thus, Eq. 8—33 (with W = 0) leads to A
0=Kf —K,‘+"AU+AEm
2 0—K, +mgdsin¢9+pkmgdcos6
which leads to d=__l_<i__—___ﬂ_——=4_3m_ mg(sin6+ ,uk cos 6) — (4.0) (9.8) (sin30°+ 0.30cos30°) ...

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- Spring '08
- Richardr
- mechanics, Work