Physics Solution Manual for 1100 and 2101

Solution setting the two expressions for work equal

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Unformatted text preview: shing force. The work done by a constant force is given by Equation 6.1 as W = ( F cos θ ) s , where F is the magnitude of the force, s is the magnitude of the displacement, and θ is the angle between the force and the displacement. SOLUTION a. The x-component of the net force is zero, ΣFx = 0, so that F cos 29.0° − f = 0 1442444 4 3 (4.9a) ΣFx The magnitude of the force that the shopper exerts is F = 48.0 N f = = 54.9 N . cos 29.0° cos 29.0° b. The work done by the pushing force F is W = ( F cos θ ) s = ( 54.9 N ) ( cos 29.0° ) ( 22.0 m ) = 1060 J (6.1) c. The angle between the frictional force and the displacement is 180°, so the work done by the frictional force f is W = ( f cos θ ) s = ( 48.0 N ) ( cos 180.0° ) ( 22.0 m ) = −1060 J d. The angle between the weight of the cart and the displacement is 90°, so the work done by the weight mg is 288 WORK AND ENERGY ( W = ( mg cos θ ) s = (16.0 kg ) 9.80 m/s 2 ) ( cos 90°) ( 22.0 m ) = 0 J ______________________________________________________________________________ 9. SSM REASONING AND SOLUTION a. According to Equation 6.1, the work done by the applied force is 3 W = Fs cos θ = (2.40 × 102 N)(8.00 m) cos 20.0° = 1.80 × 10 J b. According to Equation 6.1, the work done by the frictional force is Wf = fks cos θ, where f k = µs (mg − F sin θ ) = (0.200) (85.0 kg)(9.80 m/s2 ) − (2.40 × 102 N)(sin 20.0°) = 1.50 × 102 N Therefore, 3 Wf = (1.50 × 10 N)(8.00 m) cos 180° = −1.20 × 10 J 2 ______________________________________________________________________________ 10. REASONING The amount of work W done by a wind force F is given by W = ( F cos θ ) s (Equation 6.1), where F is the magnitude of the force, θ is the angle between the direction of the force and the direction of the boat’s motion (which is due north), and s is the magnitude of the boat’s displacement. In the two scenarios described, the magnitudes of both wind forces F1 and F2 are the same and equal F, but the displacements and angles are different. Therefore, we can express the work done in the first case as W1 = ( F cos θ1 ) s1 and that done in the second case as W2 = ( F cos θ 2 ) s2 . We will use these two expressions, along with the fact that the wind does the same amount of work in both cases ( W1 = W2 ) to find the angle θ1 between due north and the direction of the wind’s force. SOLUTION Setting the two expressions for work done by the wind equal, and solving for the angle θ1, we find F cos θ1 ) s F cos θ 2 ) s (142441 = (14 2442 4 3 4 3 W1 W2 or cos θ1 = s2 cos θ 2 s1 or s2 cos θ 2 s1 θ1 = cos −1 In the second case, the force of the wind points due north, the same direction that the boat sails. Therefore, the second angle is zero (θ2 = 0°), and angle that the force of the wind makes with due north is Chapter 6 Problems θ1 = cos ( 289 ) ( 47 m ) cos 0o = 25o 52 m −1 11. REASONING The work done by a constant force of magnitude F that makes an angle θ with a displacement of magnitude s is given by W = ( F cos θ ) s (Equation 6.1). The magnitudes and directions of the pulling forces exerted by the husband and wife differ, but the displacement of the wagon is the same in both cases. Therefore, we can express the work WH done by the husband’s pulling force as WH = ( FH cos θ H ) s . Similarly, the work done by the wife can be written as WW = ( FW cos θ W ) s . We know the directional angles of both forces, the magnitude FH of the husband’s pulling force, and that both forces do the same amount of work (WW = WH). SOLUTION Setting the two expressions for work equal to one another, and solving for the magnitude FW of the wife’s pulling force, we find FW cos θ W ) FH s (144244s = (14cos θ H )3 4 244 3 WW or FW = WH FH cos θ H cos θ W = ( 67 N ) ( cos 58o ) cos 38o = 45 N 12. REASONING AND SOLUTION The net work done on the car is WT = WF + Wf + Wg + WN WT = Fs cos 0.0° + f s cos 180° – mgs sin 5.0° + FNs cos 90° Rearranging this result gives WT F= + f + mg sin 5.0° s ( ) 150 × 103 J 2 3 + 524 N + (1200 kg ) 9.80 m/s sin 5.0° = 2.07 × 10 N 290 m ______________________________________________________________________________ = 290 WORK AND ENERGY 13. REASONING The work W done by the net external force acting on the skier can be found 2 from the work-energy theorem in the form of W = 1 mvf2 − 1 mv0 (Equation 6.3) because 2 2 the final and initial kinetic energies of the skier can be determined. The kinetic energy ( mv ) 1 2 2 is increasing, because the skier’s speed is increasing. Thus, the work will be positive, reflecting the fact that the net external force must be in the same direction as the displacement of the skier to make the skier pick up speed. SOLUTION The work done by the net external force acting on the skier is W = 1 mvf2 − 2 = 1 2 1 mv 2 0 2 ( 70.3 kg )(11.3 m /s )2 − 1 ( 70.3 kg )( 6.10 m /s )2 = 2 3 3.2 × 10 J ______________________________________________________________________________ 14. REASONING The car’s kinetic energy depends upon its mass and speed via KE = 1 mv 2 2 (Equation 6.2). The total amount of work done on the car is equal to the difference between its final and initial kinetic energies: W = KE f −...
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