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Physics Solution Manual for 1100 and 2101

# Substituting this result into equation 1 gives fs mghf

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Unformatted text preview: of the run is 2 2 vf = v0 + 2 g ( h0 − hf ) = ( 0 m/s ) + 2 ( 9.80 m/s 2 ) (104 m ) = 45.1 m/s b. The work Wnc done by the nonconservative forces follows directly from Equation 6.6: ( ) 2 Wnc = 1 m vf2 − v0 + mg ( hf − h0 ) 2 = 1 2 ( 86.0 kg ) ( 35.8 m/s )2 − ( 0 m/s )2 + (86.0 kg ) ( 9.80 m/s 2 ) ( −104 m ) = −3.25 ×104 J Note that the difference in heights, hf − h0 = −104 m , is a negative number because the final height hf is less than the initial height h0. ______________________________________________________________________________ 57. REASONING a. Because the kinetic frictional force, a nonconservative force, is present, it does negative work on the skier. The work-energy theorem, in the form of Equation 6.6, may be used to find the work done by this force. b. Once the work done by the kinetic frictional force is known, the magnitude of the kinetic frictional force can be determined by using the definition of work, Equation 6.1, since the magnitude of the skier’s displacement is known. 320 WORK AND ENERGY SOLUTION a. The work Wnc done by the kinetic frictional force is related to the object’s kinetic and potential energies by Equation 6.6: Wnc = ( 1 2 ) + ( mgh 2 mvf2 − 1 mv0 2 − mgh0 ) f s The initial height of the skier at the bottom of the hill is h0 = 0 m, and the final height is hf = s sin 25° (see the drawing). Thus, the work is Wnc = ( = 1 2 1 mv 2 f 2 2 − 1 mv0 2 25° ) + mg ( s sin 25° − h ) 0 ( 63 kg )( 4.4 m / s )2 − 1 ( 63 kg )( 6.6 m / s )2 2 ( + ( 63 kg ) 9.80 m / s 2 ) (1.9 m ) sin 25° − 0 m = −270 J b. The work done by the kinetic frictional force is, according to Equation 6.1, Wnc = ( f k cos 180° ) s , where fk is the magnitude of the kinetic frictional force, and s is the magnitude of the skier’s displacement. The displacement of the skier is up the hill and the kinetic frictional force is directed down the hill, so the angle between the two vectors is θ = 180° and cos θ = –1. Solving the equation above for fk, we have Wnc −270 J = = 140 N − s −1.9 m ______________________________________________________________________________ fk = 58. REASONING AND SOLUTION Equation 6.8, we have Wnc = ( According to the work-energy theorem as given in 1 mvf2 2 + mghf )−( 1 2 mv0 2 + mgh0 ) The metal piece starts at rest and is at rest just as it barely strikes the bell, so that ( ) vf = v0 = 0 m/s. In addition, hf = h and h0 = 0 m, while Wnc = 0.25 1 M v , where M and 2 v are the mass and speed of the hammer. Thus, the work-energy theorem becomes 0.25 ( 1 M v2 2 Solving for the speed of the hammer, we find ) = mgh 2 hf Chapter 6 Problems 321 2mgh 2(0.400 kg)(9.80 m/s2 )(5.00 m) = = 4.17 m/s 0.25 M 0.25 (9.00 kg) ______________________________________________________________________________ v= 59. REASONING We can determine the force exerted on the diver by the water if we first find the work done by the force. This is because the work can be expressed using Equation 6.1 as Wnc = (F cos θ)s, where F is the magnitude of the force from the water and s is the magnitude of the displacement in the water. Since the force that the water exerts is nonconservative, we have included the subscript “nc” in labeling the work. To calculate the work we will use the work-energy theorem in the form of Equation 6.8: 2 Wnc = ( 1 mvf2 + mghf ) − ( 1 mv0 + mgh0 ) . In this theorem Wnc is the work done by the net 2 2 nonconservative force, which, in this case, is just the force exerted by the water. SOLUTION We write Equation 6.8 as follows: Wnc = mv + mgh (144244 ) 3 1 2 2 f f Final mechanical energy − mv + mgh (144244 ) 3 1 2 2 0 0 = mghf − mgh0 (1) Initial mechanical energy where we have used the fact that the diver is at rest initially and finally, so v0 = vf = 0 m/s. According to Equation 6.1 the work done on the diver by the force of the water is Wnc = (F cos θ)s = (F cos 180º)s = −Fs The angle θ between the force of the water and the displacement is 180º, because the force opposes the motion. Substituting this result into Equation (1) gives − Fs = mghf − mgh0 (2) Identifying the final position under the water as hf = 0 m und using upward as the positive direction, we know that the initial position on the tower must be h0 = 3.00 m + 1.10 m = 4.10 m. Thus, solving Equation (2) for F, we find that F= mg ( h0 − hf ) = ( 67.0 kg ) ( 9.80 m/s 2 ) ( 4.10 m ) − ( 0 m ) = 2450 N s 1.10 m ______________________________________________________________________________ 60. REASONING AND SOLUTION The force exerted by the bat on the ball is the only nonconservative force acting. The work due to this force is 2 Wnc = mvf2 – mv0 + mg ( hf – h0 ) 1 1 2 2 322 WORK AND ENERGY Taking h0 = 0 m at the level of the bat, v0 = 40.0 m/s just before the bat strikes the ball and vf to be the speed of the ball at hf = 25.0 m, we have vf = 2Wnc m 2 + v0 – 2 ghf 2(70.0 J) 2 2 + ( 40.0 m/s ) –2(9.80 m/s )(25.0 m) = 45.9 m/s 0.140 kg _________________________________________________________...
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