Unformatted text preview: HOMEWORK 13
CH 13: 4,8,10,12,20,21,25,27,36,39,51,53
13.4 (a) The spring constant is k Fs x mg 50 N 1.0 103 N m x 5.0 102 m F Fs kx 1.0 103 N m 0.11 m 1.1 102 N
(b) 13.8 (a) The graph will be a straight line passing through the origin with a slope equal to k 1.0 103 N m . When the gun is fired, the elastic potential energy initially stored in the spring is transformed into kinetic energy of the projectile. Thus, it is necessary to have 12 1 2 kx mv0 20 2 or k 2 3.00 10 3 kg 45.0 m s mv0 2 2 x0 8.00 10 2 m 2 949 N m (b) The magnitude of the force required to compress the spring 8.00 cm and load the gun is Fs k x 949 N m 8.00 102 m 75.9 N
Fmax 230 N 575 N m xmax 0.400 m 13.10 (a) k (b) work done PEs 12 1 2 kx 575 N m 0.400 46.0 J 2 2 13.12 Conservation of mechanical energy, (KE PEg PEs ) f (KE PEg PEs )i , gives
1 2 mvi2 0 0 0 0 1 2 kx 2 , or f vi k x mi 5.00 106 N m 3.16 102 m 2.23 m s 1000 kg 13.20 v k A2 x 2 m 10.0 N m 2 2 50.0 103 kg 0.250 m 0.125 m 3.06 m s 13.21 (a) The motion is simple harmonic because the tire is rotating with constant velocity and you are looking at the uniform circular motion of the “bump” projected on a p lane perpendicular to the tire. Page 13.1 Chapter 13 (b) Note that the tangential speed of a point on the rim of a rolling tire is the same as the translational speed of the axle. Thus, vt vcar 3.00 m s and the angular velocity of the tire is vt 3.00 m s 10.0 rad s r 0.300 m Therefore, the period of the motion is T 2 2 0.628 s 10.0 rad s 13.25 The spring constant is found from k 0.010 kg 9.80 m s2 Fs mg x x 3.9 102 m 2 .5 N m When the object attached to the spring has mass m = 25 g, the period of oscillation is T 2 m 2 k 0.025 kg 0.63 s 2 .5 N m 13.27 (a) The period of oscillation is T 2 m k where k is the spring constant and m is the mass of the object attached to the end of the spring. Hence, T 2 0.250 kg 1.0 s 9.5 N m (b) If the cart is released from rest when it is 4.5 cm from the equilibrium position, the amplitude of oscillation will be A = 4.5 cm = 4.5 102 m. The maximum speed is then given by vmax A A k 4.5 102 m m 9.5 N m 0.28 m s 0.250 kg (c) When the cart is 14 cm from the left end of the track, it has a displa cement of x = 14 cm 12 cm = 2.0 cm from the equilibrium position. The speed of the cart at this distance from equilibrium is v k A2 x 2 m 9.5 N m 2 2 0.045 m 0.020 m 0.25 m s 0.250 kg Page 13.2 Chapter 13 13.36 The period in Tokyo is TT 2 LT / gT and the period in Cambridge is TC 2 LC / gC . We know that TT = TC = 2.000 s, from which, we see that g L L LT 0.994 2 1.001 5 C , or C C gT LT 0.992 7 gT gC
13.39 (a) From T 2 L g , the length of a pendulum with period T is L = gT2/42. On Earth, with T = 1.0 s, L 9.80 m s2 1.0 s 4 2 2 0.25 m 25 cm If T = 1.0 s on Mars, L 3.7 m s2 1.0 s 4 2 2 0.094 m 9.4 cm (b) The period of an object on a spring is T 2 m k , which is independent of the local freefall acceleration. Thus, the same mass will work on Earth and on Mars. This mass is m 10 N m 1.0 s k T2 2 4 4 2 2 0.25 kg 13.51 (a) The speed of transverse waves in the cord is v F , where m L is the mass per unit length. With the tension in the cord being F = 12.0 N, the wave speed is v F F mL FL m 12.0 N 6.30 m 0.150 kg 22.4 m s (b) The time to travel the length of the cord is t L 6.30 m 0.281 s v 22.4 m s 13.53 (a) The mass per unit length is m 0.0600 kg 0.0120 kg m L 5.00 m
F , the required tension in the string is From v Chapter 13 F v2 50.0 m s F 2 0.0120 kg m 30.0 N , (b) v 8.00 N 25.8 m s 0.0120 kg m Page 13.4 ...
View
Full Document
 Winter '11
 ArmenN.Kocharian
 Physics, Energy, Mass, Work, 0.250 M, 0.11 M, 0.300 m

Click to edit the document details