ch7 - 7.3.IDENTIFY: Use the free-body diagram for the bag...

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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 φ = and . The free-body diagram is given in Figure 7.3b. 34.85 = ° F G is the horizontal force applied by the worker. In the calculation of take grav U y + upward and 0 y = at the initial position of the bag. EXECUTE: (a) gives 0 y F = 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) so . tot 0 W = 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 directly. worker W Figure 7.3 7.5. IDENTIFY and SET UP: Use energy methods. (a) 1 1 other 2 2 . K UW KU ++ = + 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 11 2 ; K mv = 2 1 22 2 K mv = , 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 2 1 2 (12.0 m/s) 2(9.80 m/s )(22.0 m) 24.0 m/s vvg y =+ = + = EVALUATE: The projection angle of 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. 53.1 ° (b) Nothing changes in the calculation. The expression derived in part (a) for is independent of the angle, so the same as in part (a). 2 v 2 24.0 m/s, v = (c) The ball travels a shorter distance in part (b), so in that case air resistance will have less effect.
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7.9.IDENTIFY: tot B A WK K =− . The forces on the rock are gravity, the normal force and friction. SET UP: Let at point B and let 0 y = y + be upward. . The work done by friction is negative; W . . The free-body diagram for the rock at point B is given in Figure 7.9. The acceleration of the rock at this point is , upward. 0.50 m A yR == 0.22 J f 0 A K = R 2 rad / av = EXECUTE: (a) (i) The normal force is perpendicular to the displacement and does zero work. (ii) . 2 grav grav, grav, (0.20 kg)(9.80 m/s )(0.50 m) 0.98 J ABA WU U m g y =−= = = (b) . tot grav 0 ( 0.22 J) 0.98 J 0.76 J nf WW W W =++ = + + = tot B A K gives 2 1 tot 2 B mv W = . tot 22 ( 0 . 7 6 J ) 2.8 m/s 0.20 kg B W v m = . (c) Gravity is constant and equal to mg . n is not constant; it is zero at A and not zero at B . Therefore, kk f n μ = is also not constant. (d) y y Fm a = applied to Figure 7.9 gives . rad nm gm a −= 2 [2.8 m/s] (0.20 kg) 9.80 m/s 5.1 N 0.50 m v nmg R ⎛⎞ =+ = + = ⎜⎟ ⎝⎠ . EVALUATE: In the absence of friction, the speed of the rock at point B would be 23 . 1 m gR = / s . As the rock slides through point B , the normal force is greater than the weight 2.0 N mg = of the rock.
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This note was uploaded on 04/29/2008 for the course PH 1110 taught by Professor Kiel during the Fall '08 term at WPI.

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ch7 - 7.3.IDENTIFY: Use the free-body diagram for the bag...

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