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YF_ISM_07

YF_ISM_07 - POTENTIAL ENERGY AND ENERGY CONSERVATION 7 7.1...

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7-1 P OTENTIAL E NERGY AND E NERGY C ONSERVATION 7.1. I DENTIFY : grav U mgy = so grav 2 1 ( ) U mg y y Δ = S ET UP : y + is upward. E XECUTE : (a) 2 5 (75 kg)(9.80 m/s )(2400 m 1500 m) 6.6 10 J U Δ = = + × (b) 2 5 (75 kg)(9.80 m/s )(1350 m 2400 m) 7.7 10 J U Δ = = − × E VALUATE : grav U increases when the altitude of the object increases. 7.2. I DENTIFY : Apply m = F a G G to the sack to find the force. cos W Fs φ = . S ET UP : The lifting force acts in the same direction as the sack’s motion, so 0 φ = ° E XECUTE : (a) For constant speed, the net force is zero, so the required force is the sack’s weight, 2 (5.00 kg)(9.80 m/s ) 49.0 N. = (b) (49.0 N) (15.0 m) 735 J W = = . This work becomes potential energy. E VALUATE : The results are independent of the speed. 7.3. I DENTIFY : Use the free-body diagram for the bag and Newton's first law to find the force the worker applies. Since the bag starts and ends at rest, 2 1 0 K K = and tot 0 W = . S ET UP : A sketch showing the initial and final positions of the bag is given in Figure 7.3a. 2.0 m sin 3.5 m φ = and 34.85 φ = ° . The free-body diagram is given in Figure 7.3b. F G is the horizontal force applied by the worker. In the calculation of grav U take y + upward and 0 y = at the initial position of the bag. E XECUTE : (a) 0 y F = gives cos T mg φ = and 0 x F = gives sin F T φ = . Combining these equations to eliminate T gives 2 tan (120 kg)(9.80 m/s )tan34.85 820 N F mg φ = = = ° . (b) (i) The tension in the rope is radial and the displacement is tangential so there is no component of T in the direction of the displacement during the motion and the tension in the rope does no work. (ii) tot 0 W = so 2 worker grav grav,2 grav,1 2 1 ( ) (120 kg)(9.80 m/s )(0.6277 m) 740 J W W U U mg y y = − = = = = . E VALUATE : The force applied by the worker varies during the motion of the bag and it would be difficult to calculate worker W directly. Figure 7.3 7.4. I DENTIFY : Only gravity does work on him from the point where he has just left the board until just before he enters the water, so Eq.(7.4) applies. S ET UP : Let point 1 be just after he leaves the board and point 2 be just before he enters the water. y + is upward and 0 y = at the water. 7

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7-2 Chapter 7 E XECUTE : (a) 1 0 K = . 2 0 y = . 1 3.25 m y = . 1 grav,1 2 grav,2 K U K U + = + gives grav,1 2 U K = and 2 1 1 2 2 mgy mv = . 2 2 1 2 2(9.80 m/s )(3.25 m) 7.98 m/s v gy = = = . (b) 1 2.50 m/s v = , 2 0 y = , 1 3.25 m y = . 1 grav,1 2 K U K + = and 2 2 1 1 1 1 2 2 2 mv mgy mv + = . 2 2 2 2 1 1 2 (2.50 m/s) 2(9.80 m/s )(3.25 m) 8.36 m/s v v gy = + = + = . (c) 1 2.5 m/s v = and 2 8.36 m/s v = , the same as in part (b). E VALUATE : Kinetic energy depends only on the speed, not on the direction of the velocity. 7.5. I DENTIFY and S ET UP : Use energy methods. (a) 1 1 other 2 2 . K U W K U + + = + Solve for 2 K and then use 2 1 2 2 2 K mv = to obtain 2 . v other 0 W = (The only force on the ball while it is in the air is gravity.) 2 1 1 1 2 ; K mv = 2 1 2 2 2 K mv = 1 1 , U mgy = 1 22.0 m y = 2 2 0, U mgy = = since 2 0 y = for our choice of coordinates. Figure 7.5 E XECUTE : 2 2 1 1 1 1 2 2 2 mv mgy mv + = 2 2 2 2 1 1 2 (12.0 m/s) 2(9.80 m/s )(22.0 m) 24.0 m/s v v gy = + = + = E VALUATE : The projection angle of 53.1 ° doesn’t enter into the calculation. The kinetic energy depends only on the magnitude of the velocity; it is independent of the direction of the velocity.
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YF_ISM_07 - POTENTIAL ENERGY AND ENERGY CONSERVATION 7 7.1...

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