2 An $m = 2.00\U{kg}$ block is attached to a spring of force constant $k
3 = 500\U{N/m}$ as shown in Active Figure 6.8 on page 164. The block is
4 pulled $A = 5.00\U{cm}$ to the right of equilibrium and released from
5 rest. Find the speed the block has as it passes through equilibrium
7 \Part{a} the horizontal surface is frictionless and
8 \Part{b} the coefficient of friction between block and surface is
10 \end{problem*} % problem 6.30
13 For both cases we will use conservation of energy. Call the point
14 where the block is released $P_0$ and the point where the block passes
15 through equilibrium $P_1$. At $P_0$, the block has spring potential
16 energy $U_{s0} = 1/2\cdot k A^2$ and no kinetic or gravitational
17 potential energy. At $P_1$, the block has kinetic energy $K_1 =
18 1/2\cdot m v^2$ and no potential energy.
21 Without friction, the energy at $P_1$ is the same as that at $P_0$
22 because there is no energy lost to friction.
26 \frac{1}{2} k A^2 &= \frac{1}{2} m v^2 \\
27 v &= A \sqrt{\frac{k}{m}}
28 = 5\U{cm} \sqrt{\frac{500\U{kg/s}^2}{2\U{kg}}}
33 With friction, part of the initial energy $P_0$ bleeds out into internal
35 The work done by friction is given by
37 W_f = \vect{F} \cdot \vect{\Delta x}
39 Because the block is sliding the whole way in, the frictional force is
40 always maxed out at the constant
42 F_f = \mu F_N = \mu mg
44 In the direction opposite to the motion.
45 So friction from the table does
47 W_f = -F_f A = -\mu mgA
49 Where the negative sign denotes the frictional force sucking energy
52 Knowing the frictional work, the velocity at the equilibrium position
55 E_0 + W_f &= U_{s0} + W_f = E_1 = K_1 \\
56 \frac{1}{2} k A^2 - \mu mgA &= \frac{1}{2} m v^2\\
57 m v^2 &= k A^2 - 2 \mu mgA \\
58 v &= \sqrt{ \frac{k}{m} A^2 - 2 \mu g A} \\
59 &= \sqrt{ \frac{500\U{kg/s}^2}{2\U{kg}} (0.05\U{m})^2
60 - 2 \cdot 0.35 \cdot 9.8\U{m/s}^2 \cdot 0.05\U{m}} \\
64 What I was doing for \Part{b} in class on Wednesday was more
65 complicated because I had misread the question. I thought it was
66 asking us to find the \emph{maximum} speed, when it just asks for the
67 speed at equilibrium. Figuring out when the maximum speed occurs
68 requires more knowledge of differential equations than you guys are