Problem Set 2 Solution

For large x the block down after traveling across the

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Unformatted text preview: ! 2.3 m , where x is the maximum compression and the given 1 1 x4 U # x $ " % , F dx " kx 2 % k 2 values have been substituted. When there is a rough floor, it* xerts a force /k mg in a direction 2 4e that opposes the block’s velocity. It therefore does an amount of work /k mgd in slowing the 1 To sketch U(x), we note that for small x, U(x) behaves like the parabola kx 2 . For large x, the block down after traveling across the floor a distance d. After 2 m of floor, the block has energy 2 mv 2 2 ! /k mgd , which now goes x 4 compressing the spring and still overcoming the friction into 1 behavior is determined by % k 2 /* on the floor, which is kx 2 2 %4 k mgx . Use of the quadratic formula gives x"! / mg k U(x) 2 mv 2 2/ mgd 0 / mg 1 %2 ! 3% 4k5 k k E0 Upon substitution of the given values, the result is ! 1.12 m. x4 x5 x1 x2 x3 2-27. 0.6 m (1) E1 E2 x E3 = 0 E4 1 E " an2 + U # x $ To lift a small mass dm of rope onto the table,mv amount of work dW " , dm- g , z0 ! z - must be 2 done on it, where z0 " 0.6 m is the height of the table. The total amount of work that needs to be For E " E0 , the motion is unbounded; the particle may be anywhere. done is the integration over all the small segments of rope, giving For E " E1 (at the maxima in U(x)) the particle is at a point of unstable equilibrium. It may 2 z0 / gz0 remain at rest where it is, but if perturbed(/lightly,0it will move away from the equilibrium. (1) W " 6 s dz ) g ( z ! z ) " 0 2 dU What we substitute / ? W L " , 0 the x 4 m - we obtain W " 0 . When is the value of E1" m e find.4 kg - v,alues ,by setting dx ! 0.18 J . 0 " kx % kx 3 * 2 x = 0, - * are the equilibrium points U # -* $ " E1 " 121212 k* % k* " k* 2 4 4 For E " E2 , the particle is either bounded and oscillates between % x2 and x2 ; or the particle comes in from -. to - x3 and returns to -.. m1 % m2 m1 % m2 At equilibrium, like in previous problem, we have 72 CHAPTER 2 G...
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