Unformatted text preview: neglected, the only force that acts on the golf
ball is the conservative gravitational force (its weight). Since the maximum height of the
trajectory and the initial speed of the ball are known, the conservation of mechanical energy
can be used to find the kinetic energy of the ball at the top of the highest point. The
conservation of mechanical energy can also be used to find the speed of the ball when it is
8.0 m below its highest point.
SOLUTION
a. The conservation of mechanical energy, Equation 6.9b, states that
2 2 1 mv + mgh = 1 mv + mgh
f
f
0
0
2
2
123
KEf
Solving this equation for the final kinetic energy, KEf, yields 308 WORK AND ENERGY KE f = 1 mv0 + mg ( h0 − hf
2
2 = 1
2 ) ( 0.0470 kg )( 52.0 m / s )2 + ( 0.0470 kg ) ( 9.80 m / s 2 ) ( 0 m − 24.6 m ) = 52.2 J b. The conservation of mechanical energy, Equation 6.9b, states that
1 mv 2 + mgh
24
14 f 244 f
3 = Ef 1 mv 2 + mgh
24
14 0244 0
3 Ef The mass m can be eliminated algebraically from this equation, since it appears as a factor
in every term. Solving for vf and noting that the final height is hf = 24.6 m – 8.0 m = 16.6 m,
we have that vf = v0 + 2 g ( h0 − hf ) =
2 ( 52.0 m / s )2 + 2 ( 9.80 m / s 2 ) ( 0 m − 16.6 m ) = 48.8 m / s ______________________________________________________________________________
42. REASONING If air resistance is ignored, the only nonconservative force that acts on the
person is the tension in the rope. However, since the tension always points perpendicular to
the circular path of the motion, it does no work, and the principle of conservation of
mechanical energy applies.
SOLUTION Let the initial position of the person be at the top of the cliff, a distance h0
above the water. Then, applying the conservation principle to path 2, we have
1
mvf2 + mghf
24
14 244
3 Ef ( at water ) = 1
2
mv0 + mgh0
24
14 244
3 (6.9b) E0 ( at release point ) Solving this relation for the speed at the release point, we have v0 = vf2 + 2 g ( hf − h0 )
But vf = 13.0 m/s, since the speed of entry is the same for either path. Therefore,
2
v0 = vf2 + 2 g ( hf − h0 ) = (13.0 m/s ) + 2 ( 9.80 m/s 2 ) ( −5.20 m ) = 8.2 m/s _____________________________________________________________________________________________ Chapter 6 Problems 43. REASONING AND SOLUTION
Since friction and air resistance are
negligible, mechanical energy is
conserved. Thus, the ball will have the
same speed at the bottom of its swing
whether it is moving toward or away
from the crane. As the drawing shows,
when the cable swings to an angle of
20.0°, the ball will rise a distance of 309 20.0º
L
L − L cos 20.0º L cos 20.0º
h0 = 0 m h = L (1 – cos 20.0°)
where L is the length of the cable. Applying the conservation of energy with h0 = 0 m gives
1
2 2
mv0 = 1 mvf2 + mgh
2 Solving for vf gives
2
vf = v0 − 2 gL (1 − cos 20.0° ) = ( 5.00 m/s ) 2 − 2 ( 9.80 m/s 2 ) (12.0 m )(1 − cos 20.0° ) = 3.29 m/s ______________________________________________________________________________
44. REASONING To find the maximum height H above the end of the track we will analyze
the projectile motion of the skateboarder after she leaves the track. For this analysis we will
use the principle of conservation of mechanical energy, which applies because friction and
air resistance are being ignored. In applying this principle to the projectile motion,
however, we will need to know the speed of the skateboarder when she leaves the track.
Therefore, we will begin by determining this speed, also using the conservation principle in
the process. Our approach, then, uses the conservation principle twice. SOLUTION Applying the conservation of mechanical energy in the form of Equation 6.9b,
we have
2
1
mvf2 + mghf = 1 mv0 + mgh0
2
2
14 244
4
3
14 244
4
3
Final mechanical energy
at end of track Initial mechanical energy
on flat part of track We designate the flat portion of the track as having a height h0 = 0 m and note from the
drawing that its end is at a height of hf = 0.40 m above the ground. Solving for the final
speed at the end of the track gives
2
vf = v0 + 2 g ( h0 − hf ) = ( 5.4 m/s ) 2 + 2 ( 9.80 m/s 2 ) ( 0 m ) − ( 0.40 m ) = 4.6 m/s 310 WORK AND ENERGY This speed now becomes the initial speed v0 = 4.6 m/s for the next application of the
conservation principle. At the maximum height of her trajectory she is traveling
horizontally with a speed vf that equals the horizontal component of her launch velocity.
Thus, for the next application of the conservation principle vf = (4.6 m/s) cos 48º. Applying
the conservation of mechanical energy again, we have
2
1
2 mvf + mghf
14 244
4
3
Final mechanical energy
at maximum height of trajectory = 2
1
2 mv0 + mgh0
14 244
4
3
Initial mechanical energy
upon leaving the track Recognizing that h0 = 0.40 m and hf = 0.40 m + H and solving for H give
1
2 H= 45. 2
mvf2 + mg ( 0.40 m ) + H = 1 mv0 + mg ( 0.40 m ) 2 2
v0 − vf2 2g = ( 4.6 m/s ) 2 − ( 4.6 m/s ) cos 48° 2 ( 9.80 m/s 2 ) 2 = 0.60 m SSM REASONING Friction and air resist...
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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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