Physics Solution Manual for 1100 and 2101

Physics Solution Manual for 1100 and 2101

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Unformatted text preview: 6.6, which states that this work is equal to the surfer’s change 2 in kinetic energy, 1 mvf2 − 1 mv0 , plus the change in her potential energy, mghf − mgh0 : 2 2 Wnc = ( 1 mvf2 − 1 mv02 ) + ( mghf − mgh0 ) 2 2 (6.6) All the terms in this equation are known, so we can evaluate the work. SOLUTION We note that the difference between her final and initial heights is hf − h0 = −2.7 m, negative because her initial height h0 is greater than her final height hf. Thus, the work done by the nonconservative force is: ( ) 2 Wnc = 1 m vf2 − v0 + mg ( hf − h0 ) 2 = 1 2 ( 59 kg ) ( 9.5 m/s )2 − (1.4 m/s )2 + ( 59 kg ) ( 9.80 m/s2 ) ( −2.7 m ) = 1.0 ×103 J ______________________________________________________________________________ 52. REASONING As the firefighter slides down the pole from a height h0 to the ground (hf = 0 m), his potential energy decreases to zero. At the same time, his kinetic energy increases as he speeds up from rest (v0 = 0 m/s) to a final speed vf,. However, the upward nonconservative force of kinetic friction fk, acting over a downward displacement h0, does a ( ) negative amount of work on him: Wnc = f k cos180o h0 = − f k h0 (Equation 6.1). This work decreases his total mechanical energy E. Applying the work-energy theorem (Equation 6.8), with hf = 0 m and v0 = 0 m/s, we obtain Wnc = ( 12 mvf2 + mghf ) − ( 1 mv02 + mgh0 ) 2 ( ) − f k h0 = 1 mvf2 + 0 J − ( 0 J + mgh0 ) = 1 mvf2 − mgh0 2 { 14 244 14243 2 4 3 Wnc Ef E0 (6.8) (1) Chapter 6 Problems 317 SOLUTION Solving Equation (1) for the height h0 gives mgh0 − f k h0 = 1 mvf2 2 h0 ( mg − f k ) = 1 mvf2 2 or or h0 = mvf2 2 ( mg − f k ) The distance h0 that the firefighter slides down the pole is, therefore, ( 93 kg )( 3.4 m/s )2 h0 = = 2 ( 93 kg ) ( 9.80 m/s 2 ) − 810 N 53. 5.3 m SSM REASONING AND SOLUTION The work-energy theorem can be used to determine the change in kinetic energy of the car according to Equation 6.8: Wnc = + mgh ) ( mv + mgh ) ( mv 2443 − 1442443 144 1 2 2 f 1 2 f 2 0 Ef 0 E0 The nonconservative forces are the forces of friction and the force due to the chain mechanism. Thus, we can rewrite Equation 6.8 as Wfriction + Wchain = ( KE f + mghf ) − ( KE 0 + mgh0 ) We will measure the heights from ground level. Solving for the change in kinetic energy of the car, we have KE f − KE 0 = Wfriction + Wchain − mghf + mgh0 = − 2.00×104 J + 3.00 × 104 J − (375 kg)(9.80 m/s 2 )(20.0 m) + (375 kg)(9.80 m/s 2 )(5.00 m) = –4.51×104 J ______________________________________________________________________________ 54. REASONING We will use the work-energy theorem Wnc = Ef − E0 (Equation 6.8) to find the speed of the student. Wnc is the work done by the kinetic frictional force and is negative because the force is directed opposite to the displacement of the student. SOLUTION The work-energy theorem states that 2 ( 2 ) Wnc = 1 mvf + mghf − 1 mv0 + mgh0 24 2 14 244 3 1442443 4 4 Ef E0 (6.8) 318 WORK AND ENERGY Solving for the final speed gives vf = 2Wnc m 2 + v0 − 2 g ( hf − h0 ) ( 3 2 −6.50 × 10 J ) + ( 0 m /s ) ( ) ( −11.8 m ) = 8.6 m / s 83.0 kg ______________________________________________________________________________ = 55. 2 − 2 9.80 m /s 2 SSM REASONING The work-energy theorem can be used to determine the net work done on the car by the nonconservative forces of friction and air resistance. The workenergy theorem is, according to Equation 6.8, Wnc = ( 1 mv 2 f 2 )( + mghf − 1 mv 2 0 2 + mgh0 ) The nonconservative forces are the force of friction, the force of air resistance, and the force provided by the engine. Thus, we can rewrite Equation 6.8 as Wfriction + Wair + Wengine = ( 1 mv 2 f 2 )( + mghf − 1 mv 2 0 2 + mgh0 ) This expression can be solved for Wfriction + Wair . SOLUTION We will measure the heights from sea level, where h0 = 0 m. Since the car starts from rest, v0 = 0 m/s. Thus, we have Wfriction + Wair = m ( 1 v2 2f ) + ghf − Wengine 3 2 2 2 6 = (1.50 × 10 kg) 1 (27.0 m/s) + (9.80 m/s )(2.00 × 10 m) – (4.70 × 10 J) 2 6 = –1.21 × 10 J ______________________________________________________________________________ 56. REASONING According to Equation 6.6, the work Wnc done by nonconservative forces, such as kinetic friction and air resistance, is equal to the object’s change in kinetic energy, 1 mv 2 − 1 mv 2 plus the change in its potential energy, mgh − mgh : f 0 f 0 2 2 Wnc = ( 1 mvf2 − 1 mv02 ) + ( mghf − mgh0 ) 2 2 This relation will be used in both parts (a) and (b) of the problem. (6.6) Chapter 6 Problems 319 SOLUTION a. Since nonconservative forces are absent, the work done by them is zero, so Wnc = 0 J. In this case, we can algebraically rearrange Equation 6.6 to show that the initial total mechanical energy (kinetic energy plus potential energy) E0 at the top is equal to the final total mechanical energy Ef at the bottom: 1 mv 2 + mgh = 1 mv 2 + mgh f 244f 02440 2 2 144 3 144 3 Ef E0 Solving for the final speed vf of the skeleton and rider gives 2 vf = v0 + 2 g ( h0 hf ) The initial speed of the rider is v0 = 0 m/s, and the drop in height is h0 − hf = 104 m , so the final speed at the bottom...
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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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