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Physics Solution Manual for 1100 and 2101

# Reasoning and solution a in both cases the lift force

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Unformatted text preview: ee-body diagram at the right shows the three forces that act on you: W is your weight, Wb is the weight of your belongings, and FN is the normal force exerted on you by the floor of the elevator. Since you are moving upward at a constant velocity, your acceleration is zero, you are in equilibrium, and the net force in the y direction must be zero: FN W Wb +y Chapter 6 Problems FN − W − Wb = 0 14 244 4 3 ΣFy 285 (4.9b) Therefore, the magnitude of the normal force is FN = W + Wb. The work done by the normal force is W = ( FN cos θ ) s = (W + Wb ) ( cos θ ) s (6.1) = ( 685 N + 915 N )( cos 0° )(15.2 m ) = 24 300 J b. During the downward trip, you are still in equilibrium since the elevator is moving with a constant velocity. The magnitude of the normal force is now FN = W. The work done by the normal force is W = ( FN cos θ ) s = (W cos θ ) s = ( 685 N ) ( cos 180° ) (15.2 m ) = −10 400 J (6.1) ______________________________________________________________________________ 3. REASONING The work done by the tension in the cable is given by Equation 6.1 as W = (T cos θ ) s . Since the elevator is moving upward at a constant velocity, it is in equilibrium, and the magnitude T of the tension must be equal to the magnitude mg of the elevator’s weight; T = mg. SOLUTION a. The tension and the displacement vectors point in the same direction (upward), so the angle between them is θ = 0°. The work done by the tension is W = (T cos θ ) s = ( mg cos θ ) s (6.1) = (1200 kg ) ( 9.80 m/s 2 ) cos 0° ( 35 m ) = 4.1 × 105 J b. The weight and the displacement vectors point in opposite directions, so the angle between them is θ = 180°. The work done by the weight is W = ( mg cos θ ) s = (1200 kg ) ( 9.80 m/s 2 ) cos 180° ( 35 m ) = −4.1 × 105 J (6.1) ______________________________________________________________________________ 4. REASONING a. The work done by the gravitational force is given by Equation 6.1 as W = (F cos θ) s. The gravitational force points downward, opposite to the upward vertical displacement of 4.60 m. Therefore, the angle θ is 180º. 286 WORK AND ENERGY b. The work done by the escalator is done by the upward normal force that the escalator exerts on the man. Since the man is moving at a constant velocity, he is in equilibrium, and the net force acting on him must be zero. This means that the normal force must balance the man’s weight. Thus, the magnitude of the normal force is FN = mg, and the work that the escalator does is also given by Equation 6.1. However, since the normal force and the upward vertical displacement point in the same direction, the angle θ is 0º. SOLUTION a. According to Equation 6.1, the work done by the gravitational force is W = ( F cos θ ) s = ( mg cos θ ) s = ( 75.0 kg ) ( 9.80 m/s 2 ) cos180° ( 4.60 m ) = −3.38 × 103 J b. The work done by the escalator is W = ( F cos θ ) s = ( FN cos θ ) s = ( 75.0 kg ) ( 9.80 m/s 2 ) cos 0° ( 4.60 m ) = 3.38 × 103 J 5. SSM REASONING AND SOLUTION Solving Equation 6.1 for the angle θ , we obtain 1.10 × 103 J W = cos −1 = 42.8° Fs (30.0 N)(50.0 m) ______________________________________________________________________________ θ = cos −1 6. REASONING AND SOLUTION According to Equation 6.1, W = Fs cos θ, the work is a. W = (94.0 N)(35.0 m) cos 25.0° = 2980 J b. W = (94.0 N)(35.0 m) cos 0° = 3290 J ______________________________________________________________________________ 7. REASONING AND SOLUTION a. In both cases, the lift force L is perpendicular to the displacement of the plane, and, therefore, does no work. As shown in the drawing in the text, when the plane is in the dive, there is a component of the weight W that points in the direction of the displacement of the plane. When the plane is climbing, there is a component of the weight that points opposite to the displacement of the plane. Thus, since the thrust T is the same for both cases, the net force in the direction of the displacement is greater for the case where the plane is diving. Since the displacement s is the same in both cases, more net work is done during the dive . Chapter 6 Problems 287 b. The work done during the dive is Wdive = (T + W cos 75°) s , while the work done during the climb is Wclimb = (T + W cos 115°) s . Therefore, the difference between the net work done during the dive and the climb is Wdive – Wclimb = ( T + W cos 75°) s – ( T + W cos 115°) s = Ws (cos 75° – cos 115°) = (5.9 × 10 4 N)(1.7 × 103 m)(cos 75° – cos 115°) = 6.8 × 10 7 J ______________________________________________________________________________ 8. REASONING The drawing shows three of the +y 29.0° forces that act on the cart: F is the pushing force that F the shopper exerts, f is the frictional force that f +x opposes the motion of the cart, and mg is its weight. The displacement s of the cart is also shown. Since the cart moves at a constant velocity along the +x mg s direction, it is in equilibrium. The net force acting on it in this direction is zero, ΣFx = 0. This relation can be used to find the magnitude of the pu...
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