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. IDENTIFY: grav Um g y = so grav 2 1 () g y y Δ= SET UP: y + is upward. EXECUTE: (a) 25 (75 kg)(9.80 m/s )(2400 m 1500 m) 6.6 10 J U = + × (b) (75 kg)(9.80 m/s )(1350 m 2400 m) 7.7 10 J U = − × EVALUATE: grav U increases when the altitude of the object increases. 7.2. IDENTIFY: Apply m = F a G G to the sack to find the force. cos WF s φ = . SET UP: The lifting force acts in the same direction as the sack’s motion, so 0 = ° EXECUTE: (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. EVALUATE: The results are independent of the speed. 7.3. IDENTIFY: 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, 21 0 KK −= and tot 0 W = . SET 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 = and 34.85 = ° . 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. EXECUTE: (a) 0 y F = gives cos Tm g = and 0 x F = gives sin FT = . Combining these equations to eliminate T gives 2 tan (120 kg)(9.80 m/s )tan34.85 820 N Fm g = ° . (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 WW U U m g y y =− = = = = . EVALUATE: 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. IDENTIFY: 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. SET 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 EXECUTE: (a) 1 0 K = . 2 0 y = . 1 3.25 m y = . 1g r a v , 1 2g r a v , 2 KU += + gives grav,1 2 UK = and 2 1 12 2 mgy mv = . 2 21 2 2(9.80 m/s )(3.25 m) 7.98 m/s vg y == = . (b) 1 2.50 m/s v = , 2 0 y = , 1 3.25 m y = . r a v , 1 2 K and 22 11 112 mv mgy mv . 2 1 2 (2.50 m/s) 2(9.80 m/s )(3.25 m) 8.36 m/s vvg y =+ = + = . (c) 1 2.5 m/s v = and 2 8.36 m/s v = , the same as in part (b). EVALUATE: Kinetic energy depends only on the speed, not on the direction of the velocity. 7.5. IDENTIFY and SET UP: Use energy methods. (a) 1 1 other 2 2 . KUW ++ = + Solve for 2 K and then use 2 1 2 Km v = to obtain 2 . v other 0 W = (The only force on the ball while it is in the air is gravity.) 2 1 2 ; v = 2 1 2 v = , Um g y = 1 22.0 m y = 0, g y since 2 0 y = for our choice of coordinates. Figure 7.5 EXECUTE: mv mgy mv 2 1 2 (12.0 m/s) 2(9.80 m/s )(22.0 m) 24.0 m/s y = + = EVALUATE: 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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This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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

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