Physics Solution Manual for 1100 and 2101

# 5 107 j 23 105 n cos 0 87 m 25 107 j 18 reasoning

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Unformatted text preview: KE 0 (Equation 6.3). We will use these two relationships to determine the car’s mass. SOLUTION Combining KE = 1 mv 2 (Equation 6.2) and W = KE f − KE 0 (Equation 6.3), 2 we obtain 2 1 m v2 − v2 = W or W = 1 mvf2 − 1 mv0 f 0 2 2 2 ( ) Solving this expression for the car’s mass m, and noting that 185 kJ = 185 000 J, we find that 2 (185 000 J ) 2W m= 2 2 = = 1450 kg vf − v0 ( 28.0 m/s )2 − ( 23.0 m/s )2 15. SSM WWW REASONING AND SOLUTION The work done on the arrow by the bow is given by W = Fs cos 0° = Fs This work is converted into kinetic energy according to the work energy theorem. 2 W = 1 mvf2 − 1 mv0 2 2 Solving for vf, we find that Chapter 6 Problems 2W 2 + v0 = m vf = 2 Fs 2 + v0 = m 2 ( 65 N )( 0.90 m ) 291 + ( 0 m/s ) = 39 m/s 2 75 × 10 kg ______________________________________________________________________________ –3 16. REASONING a. Because we know the flea’s mass m, we can find its speed vf once we know its kinetic energy KE f = 1 mvf2 (Equation 6.2) at the instant it leaves the ground. The work-energy 2 theorem W = KE f − KE 0 (Equation 6.3) relates KEf to the total work W done on the flea during push-off. Because it starts from rest, the flea has no initial kinetic energy, and we see that the total work done on the flea is equal to its final kinetic energy: W = KE f − ( 0 J ) = KE f = 1 mvf2 . Since air resistance and the flea’s weight are being 2 ignored, the only force doing work on the flea is the upward force Fground exerted by the ground, so Wground = W = 1 mvf2 . 2 ( ) b. We will make use of the definition of work Wground = Fground cos θ s (Equation 6.1) to calculate the flea’s upward displacement s while the force Fground is doing work on the flea. SOLUTION a. Solving Wground = 1 mvf2 for the final speed vf of the flea, we find 2 1 2 mvf2 = Wground vf2 = or 2Wground or m vf = 2Wground m Therefore, vf = ( 2 +2.4 ×10−4 J 1.9 ×10 −4 kg ) = 1.6 m/s b. The force Fground of the ground on the flea is upward, the same as the direction of the flea’s displacement s while it is pushing off. Therefore, the angle θ between Fground and s is ( ) zero. Solving Wground = Fground cos θ s for the displacement magnitude s, we obtain s= Wground Fground cos θ = +2.4 × 10 −4 J ( 0.38 N ) ( cos 0 o ) = 6.3 × 10−4 m 292 WORK AND ENERGY 17. REASONING The work done by the catapult Wcatapult is one contribution to the work done by the net external force that changes the kinetic energy of the plane. The other contribution is the work done by the thrust force of the plane’s engines Wthrust. According to the workenergy theorem (Equation 6.3), the work done by the net external force Wcatapult + Wthrust is equal to the change in the kinetic energy. The change in the kinetic energy is the given kinetic energy of 4.5 × 107 J at lift-off minus the initial kinetic energy, which is zero since the plane starts at rest. The work done by the thrust force can be determined from Equation 6.1 [W = (F cos θ) s], since the magnitude F of the thrust is 2.3 × 105 N and the magnitude s of the displacement is 87 m. We note that the angle θ between the thrust and the displacement is 0º, because they have the same direction. In summary, we will calculate Wcatapult from Wcatapult + Wthrust = KEf − KE0. SOLUTION According to the work-energy theorem, we have Wcatapult + Wthrust = KEf − KE0 Using Equation 6.1 and noting that KE0 = 0 J, we can write the work energy theorem as follows: Wcatapult + ( F cos θ ) s = KE f 14 3 24 Work done by thrust Solving for Wcatapult gives Wcatapult = KE f − ( F cosθ ) s 14 3 24 Work done by thrust = 4.5 ×107 J − ( 2.3 ×105 N ) cos 0° ( 87 m ) = 2.5 ×107 J 18. REASONING AND SOLUTION From the work-energy theorem, Equation 6.3, 1 1 1 2 2 W = mvf2 − mv0 = m vf2 − v0 2 2 2 1 2 2 a. W = (7420 kg) (8450 m/s) – (2820 m/s) = 2 2.35 × 1011 J b. W = (7420 kg) (2820 m/s) − (8450 m/s) = –2.35 × 10 J 2 ______________________________________________________________________________ 1 2 2 11 Chapter 6 Problems 19. 293 SSM REASONING The work done to launch either object can be found from 1 1 2 2 Equation 6.3, the work-energy theorem, W = KEf − KE 0 = mvf − mv0 . 2 2 SOLUTION a. The work required to launch the hammer is ( ) 2 2 W = 1 mvf2 − 1 mv0 = 1 m vf2 − v0 = 1 (7.3 kg) (29 m/s)2 − ( 0 m/s ) = 3.1×103 J 2 2 2 2 2 b. Similarly, the work required to launch the bullet is ( ) 2 W = 1 m vf2 − v0 = 1 (0.0026 kg) (410 m/s) 2 − ( 0 m/s ) = 2.2 × 102 J 2 2 2 ______________________________________________________________________________ 20. REASONING a. The work W done by the force is related to the change in the asteroid’s kinetic energy by 2 the work-energy theorem W = 1 mvf2 − 1 mv0 (Equation 6.3). Since the final speed of the 2 2 asteroid is vf = 5500 m/s and is less than the initial speed of v0 = 7100 m/s, the work obtained from Equation 6.3 will be negative. This is consistent with the fact that, since the asteroid is slowing down, it is decelerating, so that the force...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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