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Unformatted text preview: at given, the net work W is the total area under the
graph. For the purpose of calculation, it is convenient to divide this area into two pieces, a
triangle (from s = 0 m to s = 10.0 m) and a rectangle (from s = 10.0 m to s = 20.0 m). Both
pieces have the same width (10.0 m) and height (10.0 N), so the triangle has half the area of
the rectangle. 2W
.
m
done on the object from s = 0 m to s = 10.0 m SOLUTION Solving W = 1 mvf2 for the final speed of the object, we obtain vf =
2
The net work W is the sum of the work W0,10 (the triangular area) and the work W10,20 done on the object from s = 10.0 m to s = 20.0 m
(the rectangular area). We calculate the work W10,20 by multiplying the width (10.0 m) and height (10.0 N) of the rectangle: W10,20 = ( Width ) ( Height ) . The triangle’s area, which is
the work W0,10, is half this amount: W0,10 = 1 W10,20 . Therefore, the net work
2 W = W0,10 + W10,20 done during the entire interval from s = 0 m to s = 20.0 m is 332 WORK AND ENERGY W = 1 W10,20 + W10,20 = 3 W10,20 =
2
2 3
2 ( Width ) ( Height ) We can now calculate the final speed of the object from vf = vf = 2 2W
:
m ( 3 ) ( Width )( Height ) = 3 (10.0 m )(10.0 N ) = 7.07 m/s
2
m 6.00 kg 76. REASONING AND SOLUTION The work done by the retarding force is given by
Equation 6.1: W = ( F cos θ )s . Since the force is a retarding force, it must point opposite to
the direction of the displacement, so that θ = 180° . Thus, we have W = ( F cosθ )s = (3.0 × 103 N)(cos 180°)(850 m) = –2.6 × 106 J
The work done by this force is negative , because the retarding force is directed opposite
to the direction of the displacement of the truck.
______________________________________________________________________________
77. SSM REASONING The only two forces that act on the gymnast are his weight and the
force exerted on his hands by the high bar. The latter is the (nonconservative) reaction
force to the force exerted on the bar by the gymnast, as predicted by Newton's third law.
This force, however, does no work because it points perpendicular to the circular path of
motion. Thus, Wnc = 0 J, and we can apply the principle of conservation of mechanical
energy.
SOLUTION The conservation principle gives
1
mv 2 + mgh
2 4f 244f
14
3
Ef 1 2
= mv 0 + mgh0
24
14 244
3
E0 Since the gymnast's speed is momentarily zero at the top of the swing, v0 = 0 m/s. If we
take hf = 0 m at the bottom of the swing, then h0 = 2 r , where r is the radius of the circular
path followed by the gymnast's waist. Making these substitutions in the above expression
and solving for vf , we obtain vf = 2 gh0 = 2 g(2r) = 2(9.80 m/s2 )(2 × 1.1 m) = 6.6 m/s
______________________________________________________________________________ Chapter 6 Problems 333 78. REASONING
a. Since there is no air friction, the only force that acts on the projectile is the conservative
gravitational force (its weight). The initial and final speeds of the ball are known, so the
conservation of mechanical energy can be used to find the maximum height that the
projectile attains.
b. When air resistance, a nonconservative force, is present, it does negative work on the
projectile and slows it down. Consequently, the projectile does not rise as high as when
there is no air resistance. The workenergy theorem, in the form of Equation 6.6, may be
used to find the work done by air friction. Then, using the definition of work, Equation 6.1,
the average force due to air resistance can be found.
SOLUTION
a. The conservation of mechanical energy, as expressed by Equation 6.9b, states that
2 2 Ef E0 1 mv + mgh = 1 mv + mgh
24
14 f244f 14 02440
3 24
3 The mass m can be eliminated algebraically from this equation since it appears as a factor in
every term. Solving for the final height hf gives
1
2 hf = ( v02 − vf2 ) + h 0 g Setting h0 = 0 m and vf = 0 m/s, the final height, in the absence of air resistance, is
2 hf = 2 vo − vf
2g = (18.0 m / s )2 − ( 0 m/s )2 ( 2 9.80 m / s 2 ) = 16.5 m b. The workenergy theorem is Wnc = ( 1 mv 2
f
2 2
− 1 mv0
2 ) + ( mgh f − mgh0 ) (6.6) where Wnc is the nonconservative work done by air resistance. According to Equation 6.1, the work can be written as Wnc = ( FR cos 180° ) s , where FR is the average force of air resistance. As the projectile moves upward, the force of air resistance is directed downward,
so the angle between the two vectors is θ = 180° and cos θ = –1. The magnitude s of the
displacement is the difference between the final and initial heights, s = hf – h0 = 11.8 m.
With these substitutions, the workenergy theorem becomes ( ) 2
− FR s = 1 m vf2 − vo + mg ( hf − h0 )
2 334 WORK AND ENERGY Solving for FR gives FR = = 1m
2 ( vf2 − vo2 ) + mg ( hf − h0 )
−s 1
2 ( 0.750 kg ) ( 0 m/s )2 − (18.0 m/s )2 + ( 0.750 kg ) ( 9.80 m/s2 ) (11.8 m ) =
− (11.8 m ) 2.9 N ______________________________________________________________________________
79. SSM REASONING AND SOLUTION We will assume that the tugofwar rope remains
parallel to the ground, so that the force that moves team B is in the same direction as the...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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