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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 workenergy 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 workenergy 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 workenergy 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 workenergy 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 workenergy 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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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