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216_PartUniversity Physics Solution

216_PartUniversity Physics Solution - Potential Energy and...

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Potential Energy and Energy Conservation 7-5 7.14. I DENTIFY : Only gravity does work, so apply Eq.(7.4). Use m = F a ! ! to calculate the tension. S ET UP : Let 0 y = at the bottom of the arc. Let point 1 be when the string makes a 45 ° angle with the vertical and point 2 be where the string is vertical. The rock moves in an arc of a circle, so it has radial acceleration 2 rad / a v r = E XECUTE : (a) At the top of the swing, when the kinetic energy is zero, the potential energy (with respect to the bottom of the circular arc) is (1 cos ), mgl θ where l is the length of the string and θ is the angle the string makes with the vertical. At the bottom of the swing, this potential energy has become kinetic energy, so 2 1 2 (1 cos ) , mgl θ mv = or 2 2 (1 cos ) 2(9 80 m/s ) (0 80 m) (1 cos45 ) 2.1 m/s v gl θ . . = = ° = . (b) At 45 ° from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the radial component of the weight, or 2 cos (0.12 kg) (9.80 m/s ) cos 45 0.83 N. mg θ = ° = (c) At the bottom of the circle, the tension is the sum of the weight and the mass times the radial acceleration, 2 2 (1 2(1 cos45 )) 1.9 N mg mv l mg + = + ° = E VALUATE : When the string passes through the vertical, the tension is greater than the weight because the acceleration is upward. 7.15. I DENTIFY : Apply 2 1 el 2 U kx = . S ET UP : kx F = , so 1 2 U Fx = ,where F is the magnitude of force required to stretch or compress the spring a distance x . E XECUTE : (a) (1 2)(800 N)(0.200 m) 80.0 J. = (b) The potential energy is proportional to the square of the compression or extension; 2 (80.0 J) (0.050 m 0.200 m) 5.0 J. = E VALUATE : We could have calculated 800 N 4000 N/m 0.200 m F k x = = = and then used 2 1 el 2 U kx = directly. 7.16. I DENTIFY : Use the information given in the problem with F kx = to find k . Then 2 1 el 2 U kx = . S ET UP : x is the amount the spring is stretched. When the weight is hung from the spring, F mg = . E XECUTE : 2 (3.15 kg)(9.80 m/s ) 2205 N/m 0.1340 m 0.1200 m F mg k x x = = = = . el 2 2(10.0 J) 0.0952 m 9.52 cm 2205 N/m U x k = ± = ± = ± = ± . The spring could be either stretched 9.52 cm or compressed 9.52 cm. If it were stretched, the total length of the spring would be 12.00 cm 9.52 cm 21.52 cm + = . If it were compressed, the total length of the spring would be 12.00 cm 9.52 cm 2.48 cm = . E VALUATE : To stretch or compress the spring 9.52 cm requires a force 210 N F kx = = . 7.17. I DENTIFY : Apply 2 1 el 2 U kx = . S ET UP : 2 1 0 0 2 U kx = . x is the distance the spring is stretched or compressed. E XECUTE : (a) (i) 0 2 x x = gives 2 2 1 1 el 0 0 0 2 2 (2 ) 4( ) 4 U k x kx U = = = . (ii) 0 / 2 x x = gives 2 2 1 1 1 el 0 0 0 2 4 2 ( / 2) ( ) /4 U k x kx U = = = . (b) (i) 0 2 U U = gives 2 2 1 1 0 2 2 2( ) kx kx = and 0 2 x x = . (ii) 0 /2 U U = gives 2 2 1 1 1 0 2 2 2 ( ) kx kx = and 0 / 2 x x = . E VALUATE : U is proportional to 2 x and x is proportional to U . 7.18. I DENTIFY : Apply Eq.(7.13). S ET UP : Initially and at the highest point, 0 v = , so 1 2 0 K K = = . other 0 W = . E XECUTE : (a) In going from rest in the slingshot°s pocket to rest at the maximum height, the potential energy stored in the rubber band is converted to gravitational potential energy; 3 2 (10 10 kg)(9.80 m/s ) (22.0 m) 2.16 J. U mgy = = × = (b) Because gravitational potential energy is proportional to mass, the larger pebble rises only 8.8 m.
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