Physics Solution Manual for 1100 and 2101

50 ms solving equation 1 for s gives s b 2 m vf2

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Unformatted text preview: skier (see Example 10 in Chapter 4), so FN = mg. Substituting this relation into Equation (2) gives v0 = vf2 − 2 µk m g ( cos180° ) s m = vf2 − 2 µ k g ( cos180° ) s Chapter 6 Problems 297 SOLUTION Since the skier comes to a halt, vf = 0 m/s. Therefore, the initial speed is 2 v0 = vf2 − 2 µk g ( cos180° ) s = ( 0 m/s ) − 2 ( 0.050 ) ( 9.80 m/s 2 ) ( cos180° ) ( 21 m ) = 4.5 m/s ____________________________________________________________________________________________ 25. SSM WWW REASONING When the satellite goes from the first to the second orbit, its kinetic energy changes. The net work that the external force must do to change the orbit can be found from the work-energy theorem: 1 2 1 2 2 W = KE f − KE 0 = mvf2 − mv0 . The speeds vf and v0 can be obtained from Equation 5.5 for the speed of a satellite in a circular orbit of radius r. Given the speeds, the work energy theorem can be used to obtain the work. SOLUTION According to Equation 5.5, v = GME r . Substituting into the work-energy theorem, we have 2 2 GM E GM E GM E m 1 1 1 mv 2 − 1 mv 2 = 1 m v 2 − v 2 = 1 m W=2 f − = − 0 f 0 2 2 2 rf r0 2 rf r0 Therefore, ( W= (6.67 ×10 ) –11 2 2 N ⋅ m /kg )(5.98 ×10 2 24 kg)(6200 kg) 1 1 11 × − = 1.4 × 10 J 6 7 7.0 × 10 m 3.3 × 10 m ______________________________________________________________________________ 26. REASONING a. The magnitude s of the displacement that occurs while the snowmobile coasts to a halt is the distance that we seek. The work W done by the net external force F acting on the snowmobile during the coasting phase is given by W = (F cos θ) s (Equation 6.1), and we can use this expression to obtain s. To do so, we will need to evaluate W. This we will do 2 with the aid of the work-energy theorem as given by W = 1 mvf2 − 1 mv0 (Equation 6.3), 2 2 since we know the mass m and the final and initial velocities vf and v0, respectively. The net external force F during the coasting phase is just the horizontal force of kinetic friction (magnitude = fk), since the drive force is shut off and the vertical forces balance (there is no vertical acceleration). This frictional force has a magnitude of 205 N. This follows because, while the snowmobile is moving at a constant 5.50 m/s, it is not accelerating, and the drive force must be balancing the frictional force, which exists both before and after the drive force is shut off. Finally, to use W = (F cos θ) s (Equation 6.1) to determine s, we will 298 WORK AND ENERGY need a value for the angle θ between the frictional force and the displacement. The frictional force points opposite to the displacement, so θ = 180°. b. By using only the work-energy theorem it is not possible to determine the time t during which the snowmobile coasts to a halt. To determine t it is necessary to use the equations of kinematics. We can use these equations, if we assume that the acceleration during the coasting phase is constant. The acceleration is determined by the force of friction (assumed constant) and the mass of the snowmobile, according to Newton’s second law. For example, we can use s = 1 ( v0 + vf ) t (Equation 2.7) to determine t once we know the 2 magnitude of the displacement s, which is the distance in which the snowmobile coasts to a halt. SOLUTION a. According to Equation 6.3, the work energy theorem, is 2 W = 1 mvf2 − 1 mv0 2 2 As discussed in the REASONING, we know that W = (fk cos θ) s. Substituting this result into the work-energy theorem gives ( f k cosθ ) s = 12 mvf2 − 12 mv02 (1) In Equation (1), fk = 205 N (see the REASONING), the angle θ between the frictional force and the displacement is θ = 180º (see the REASONING), the final speed is vf = 0 m/s (the snowmobile coasts to a halt), and the initial speed is given as v0 = 5.50 m/s. Solving Equation (1) for s gives s= b. 2 m ( vf2 − v0 ) (136 kg ) ( 0 m/s ) − ( 5.50 m/s ) = 2 f k cos θ 2 ( 205 N ) cos180° 2 As explained in the REASONING, we can use s = 2 = 10.0 m 1 2 (v 0 + vf ) t (Equation 2.7) to determine the time t during which the snowmobile coasts to a halt. Solving this equation for t gives 2 (10.0 m ) 2s t= = = 3.64 s v0 + vf 5.50 m/s + 0 m/s 27. REASONING AND SOLUTION The net work done on the plane can be found from the work-energy theorem: ( 2 2 W = 1 mvf2 − 1 mv0 = 1 m vf2 − v0 2 2 2 ) Chapter 6 Problems 299 The tension in the guideline provides the centripetal force required to keep the plane moving in a circular path. Since the tension in the guideline becomes four times greater, Tf = 4 T0 or mvf2 rf =4 2 mv0 r0 Solving for vf2 gives 2 vf = 2 4v0 rf r0 Substitution of this expression into the work-energy theorem gives W= 4v 2 r 1 m 2 0f r0 2 − v0 = 1 mv 2 0 2 rf 4 r 0 − 1 Therefore, 14 m 1 2 2 W = (0.90 kg)(22 m/s) 4 − 1 = 5.4 ×10 J 2 16 m ______________________________________________________________________________ 28. REASONING It is useful to divide this problem into two parts. The first part involves the skier moving on the snow. We can use the work-energy theorem to find her speed when she comes to the...
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