YF_ISM_06

YF_ISM_06 - WORK AND KINETIC ENERGY 6 6.1. 6.2. IDENTIFY:...

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6-1 W ORK AND K INETIC E NERGY 6.1. IDENTIFY: Apply Eq.(6.2). SET UP: The bucket rises slowly, so the tension in the rope may be taken to be the bucket’s weight. EXECUTE: (a) 2 (6.75 kg) (9.80 m/s )(4.00 m) 265 J. WF sm g s == = = (b) Gravity is directed opposite to the direction of the bucket’s motion, so Eq.(6.2) gives the negative of the result of part (a), or 265 J . (c) The total work done on the bucket is zero. EVALUATE: When the force is in the direction of the displacement, the force does positive work. When the force is directed opposite to the displacement, the force does negative work. 6.2. IDENTIFY: In each case the forces are constant and the displacement is along a straight line, so cos s φ = . SET UP: In part (a), when the cable pulls horizontally 0 = ° and when it pulls at 35.0 ° above the horizontal 35.0 = ° . In part (b), if the cable pulls horizontally 180 = ° . If the cable pulls on the car at 35.0 ° above the horizontal it pulls on the truck at 35.0 ° below the horizontal and 145.0 = ° . For the gravity force 90 = ° , since the force is vertical and the displacement is horizontal. EXECUTE: (a) When the cable is horizontal, 36 (850 N)(5.00 10 m)cos0 4.25 10 J W =×= × ° . When the cable is 35.0 ° above the horizontal, (850 N)(5.00 10 m)cos35.0 3.48 10 J W = × ° . (b) cos180 cos0 =− °° and cos145.0 cos35.0 , so the answers are 6 4.26 10 J −× and 6 . (c) Since cos cos90 0 ° , 0 W = in both cases. EVALUATE: If the car and truck are taken together as the system, the tension in the cable does no net work. 6.3. IDENTIFY: Each force can be used in the relation (c o s) s F s & for parts (b) through (d). For part (e), apply the net work relation as net worker grav . nf WW W W W =+ + + SET UP: In order to move the crate at constant velocity, the worker must apply a force that equals the force of friction, worker k k . Ff n μ EXECUTE: (a) The magnitude of the force the worker must apply is: () 2 worker k k k 0.25 30.0 kg 9.80 m/s 74 N n m g μμ = = = (b) Since the force applied by the worker is horizontal and in the direction of the displacement, 0 = ° and the work is: ( ) [ ] worker worker cos 74 N cos0 4.5 m 333 J s = + ° (c) Friction acts in the direction opposite of motion, thus 180 = ° and the work of friction is: ( ) [ ] k cos 74 N cos180 4.5 m 333 J f Wf s = ° (d) Both gravity and the normal force act perpendicular to the direction of displacement. Thus, neither force does any work on the crate and grav 0.0 J. n (e) Substituting into the net work relation, the net work done on the crate is: net worker grav 333 J 0.0 J 0.0 J 333 J 0.0 J W W W + + = + + + = EVALUATE: The net work done on the crate is zero because the two contributing forces, worker and , f FF are equal in magnitude and opposite in direction. 6.4. IDENTIFY: The forces are constant so Eq.(6.2) can be used to calculate the work. Constant speed implies 0. a = We must use m = F a G G applied to the crate to find the forces acting on it. 6
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6-2 Chapter 6 (a) SET UP: The free-body diagram for the crate is given in Figure 6.4.
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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_06 - WORK AND KINETIC ENERGY 6 6.1. 6.2. IDENTIFY:...

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