Physics Solution Manual for 1100 and 2101

# This work can be expressed using equation 61 as wnc

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Unformatted text preview: displacement. According to Equation 6.1, the work done by team A is W = ( F cosθ )s = (1100 N)(cos 0°)(2.0 m) = 2.2 × 10 3 J ______________________________________________________________________________ 80. REASONING AND SOLUTION The vertical height of the skier is h = s sin 14.6° so s ∆PE = mgs sin 14.6° h 14.6° = (75.0 kg)(9.80 m/s2)(2830 m) sin 14.6° = 5.24 × 105 J ______________________________________________________________________________ 81. REASONING The net work done by the pushing force and the frictional force is zero, and our solution is focused on this fact. Thus, we express this net work as WP + Wf = 0, where WP is the work done by the pushing force and Wf is the work done by the frictional force. We will substitute for each individual work using Equation 6.1 [W = (F cos θ) s] and solve the resulting equation for the magnitude P of the pushing force. SOLUTION According to Equation 6.1, the work done by the pushing force is WP = (P cos 30.0°) s = 0.866 P s The frictional force opposes the motion, so the angle between the force and the displacement is 180°. Thus, the work done by the frictional force is Wf = (fk cos 180°) s = – fk s Chapter 6 Problems Equation 4.8 indicates that the magnitude of the kinetic frictional force is fk = µ kFN, where FN is the magnitude of the normal force acting on the crate. The free-body diagram shows the forces acting on the crate. Since there is no acceleration in the vertical direction, the y component of the net force must be zero: 335 +y FN P fk 30.0º +x mg FN − mg − P sin 30.0° = 0 Therefore, FN = mg + P sin 30.0° It follows, then, that the magnitude of the frictional force is fk = µ k FN = µ k (mg + P sin 30.0°) Suppressing the units, we find that the work done by the frictional force is Wf = – fk s = − (0.200)[(1.00 × 102 kg)(9.80 m/s2) + 0.500P]s = −(0.100P + 196)s Since the net work is zero, we have WP + Wf = 0.866 Ps − (0.100P + 196)s = 0 Eliminating s algebraically and solving for P gives P = 256 N . 82. REASONING The average power generated by the tension in the cable is equal to the work done by the tension divided by the elapsed time (Equation 6.10a). The work can be related to the change in the lift’s kinetic and potential energies via the work-energy theorem. SOLUTION a. The average power is P= Wnc t (6.10a) where Wnc is the work done by the nonconservative tension force. This work is related to the lift’s kinetic and potential energies by Equation 6.6, Wnc = so the average power is P= Wnc t ( = 1 mv 2 f 2 2 − 1 mv0 2 ) + ( mgh f t ( 1 2 ) 2 mvf2 − 1 mv0 + ( mghf − mgh0 ) , 2 − mgh0 ) 336 WORK AND ENERGY Since the lift moves at a constant speed, v0 = vf, the average power becomes 83. ( 4 skiers ) 65 ( ) kg 2 9.80 m / s (140 m ) 3 skier P= = = 3.0 × 10 W t 60 s ( 2 min ) 1 min ______________________________________________________________________________ mg ( hf − h0 ) SSM REASONING AND SOLUTION a. The lost mechanical energy is Elost = E0 – Ef The ball is dropped from rest, so its initial energy is purely potential. The ball is momentarily at rest at the highest point in its rebound, so its final energy is also purely potential. Then Elost = mgh0 – mghf = ( 0.60 kg ) ( 9.80 m/s 2 ) (1.05 m ) – ( 0.57 m ) = 2.8 J b. The work done by the player must compensate for this loss of energy. Elost = ( F cos θ ) s 14 3 24 work done by player ⇒ F= Elost ( cos θ ) s = 2.8 J = 35 N ( cos 0° )( 0.080 m ) ______________________________________________________________________________ 84. REASONING When launched with the minimum initial speed v0, the puck has just enough initial kinetic energy to reach the teammate with a final speed vf of zero. All of the initial kinetic energy has served the purpose of compensating for the work done by friction in opposing the motion. Friction is a nonconservative force, so to determine v0 we will use the 2 work-energy theorem in the form of Equation 6.8: Wnc = ( 1 mvf2 + mghf ) − ( 1 mv0 + mgh0 ) . 2 2 Wnc is the work done by the net nonconservative force, which, in this case, is just the kinetic frictional force. This work can be expressed using Equation 6.1 as Wnc = (fk cos θ)s, where fk is the magnitude of the kinetic frictional force and s is the magnitude of the displacement, or the distance between the players. We are given neither the frictional force nor the distance between the players, but we are given the initial speed that enables the puck to travel half way. This information will be used to evaluate Wnc. SOLUTION Noting that a hockey rink is flat (hf = h0), we write Equation 6.8 as follows Chapter 6 Problems Wnc = mv + mgh (144244 ) 3 1 2 2 f f Final mechanical energy mv + mgh (144244 ) 3 − 2 0 1 2 2 2 = 1 mvf2 − 1 mv0 = − 1 mv0 2 2 2 0 337 (1) Initial mechanical energy where we have used the fact that the final speed is vf = 0 m/s when the initial speed v0 has its minimum value. According to Equation 6.1 the work done on the puck by the kinetic frictional force is Wnc = (fk cos θ)s = (fk cos 180º)s = − fks In this expression the angle θ is 180º, because the frictional force points opposite to the displacement. With this substitution Equation (1) becomes 2 − f k s = − 1 mv0 2 or 2 f k s = 1 mv0 2 (2) To evaluate the term fks we note that Equation (2) also applies to the failed attempt at the pass, so that f k ( 1 s ) = 1 m (1.7 m/s ) 2 2 2 or f k s = m...
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