Unformatted text preview: skier (see Example 10 in Chapter 4), so FN = mg. Substituting this relation
into Equation (2) gives
v0 = vf2 − 2 µk m g ( cos180° ) s
m = vf2 − 2 µ k g ( cos180° ) s Chapter 6 Problems 297 SOLUTION Since the skier comes to a halt, vf = 0 m/s. Therefore, the initial speed is
2
v0 = vf2 − 2 µk g ( cos180° ) s = ( 0 m/s ) − 2 ( 0.050 ) ( 9.80 m/s 2 ) ( cos180° ) ( 21 m ) = 4.5 m/s ____________________________________________________________________________________________ 25. SSM WWW REASONING When the satellite goes from the first to the second orbit,
its kinetic energy changes. The net work that the external force must do to change the orbit
can be found from the workenergy theorem: 1
2 1
2 2
W = KE f − KE 0 = mvf2 − mv0 . The speeds vf and v0 can be obtained from Equation 5.5 for the speed of a satellite in a circular
orbit of radius r. Given the speeds, the work energy theorem can be used to obtain the
work.
SOLUTION According to Equation 5.5, v = GME r . Substituting into the workenergy
theorem, we have
2
2 GM E GM E GM E m 1 1 1 mv 2 − 1 mv 2 = 1 m v 2 − v 2 = 1 m W=2 f −
=
−
0
f
0
2
2 2 rf r0 2 rf r0 Therefore, ( W= (6.67 ×10 ) –11 2 2 N ⋅ m /kg )(5.98 ×10
2 24 kg)(6200 kg) 1
1 11
×
− = 1.4 × 10 J
6
7 7.0 × 10 m 3.3 × 10 m ______________________________________________________________________________
26. REASONING
a. The magnitude s of the displacement that occurs while the snowmobile coasts to a halt is
the distance that we seek. The work W done by the net external force F acting on the
snowmobile during the coasting phase is given by W = (F cos θ) s (Equation 6.1), and we
can use this expression to obtain s. To do so, we will need to evaluate W. This we will do
2
with the aid of the workenergy theorem as given by W = 1 mvf2 − 1 mv0 (Equation 6.3),
2
2
since we know the mass m and the final and initial velocities vf and v0, respectively. The net
external force F during the coasting phase is just the horizontal force of kinetic friction
(magnitude = fk), since the drive force is shut off and the vertical forces balance (there is no
vertical acceleration). This frictional force has a magnitude of 205 N. This follows
because, while the snowmobile is moving at a constant 5.50 m/s, it is not accelerating, and
the drive force must be balancing the frictional force, which exists both before and after the
drive force is shut off. Finally, to use W = (F cos θ) s (Equation 6.1) to determine s, we will 298 WORK AND ENERGY need a value for the angle θ between the frictional force and the displacement. The
frictional force points opposite to the displacement, so θ = 180°.
b. By using only the workenergy theorem it is not possible to determine the time t during
which the snowmobile coasts to a halt. To determine t it is necessary to use the equations of
kinematics. We can use these equations, if we assume that the acceleration during the
coasting phase is constant. The acceleration is determined by the force of friction (assumed
constant) and the mass of the snowmobile, according to Newton’s second law. For
example, we can use s = 1 ( v0 + vf ) t (Equation 2.7) to determine t once we know the
2
magnitude of the displacement s, which is the distance in which the snowmobile coasts to a
halt.
SOLUTION
a. According to Equation 6.3, the work energy theorem, is
2
W = 1 mvf2 − 1 mv0
2
2 As discussed in the REASONING, we know that W = (fk cos θ) s. Substituting this result
into the workenergy theorem gives ( f k cosθ ) s = 12 mvf2 − 12 mv02 (1) In Equation (1), fk = 205 N (see the REASONING), the angle θ between the frictional force
and the displacement is θ = 180º (see the REASONING), the final speed is vf = 0 m/s (the
snowmobile coasts to a halt), and the initial speed is given as v0 = 5.50 m/s. Solving
Equation (1) for s gives s= b. 2
m ( vf2 − v0 ) (136 kg ) ( 0 m/s ) − ( 5.50 m/s ) =
2 f k cos θ
2 ( 205 N ) cos180°
2 As explained in the REASONING, we can use s = 2 = 10.0 m 1
2 (v 0 + vf ) t (Equation 2.7) to determine the time t during which the snowmobile coasts to a halt. Solving this equation for
t gives
2 (10.0 m )
2s
t=
=
= 3.64 s
v0 + vf 5.50 m/s + 0 m/s 27. REASONING AND SOLUTION The net work done on the plane can be found from the
workenergy theorem: ( 2
2
W = 1 mvf2 − 1 mv0 = 1 m vf2 − v0
2
2
2 ) Chapter 6 Problems 299 The tension in the guideline provides the centripetal force required to keep the plane moving
in a circular path. Since the tension in the guideline becomes four times greater,
Tf = 4 T0
or
mvf2
rf =4 2
mv0 r0 Solving for vf2 gives
2
vf = 2
4v0 rf r0 Substitution of this expression into the workenergy theorem gives W= 4v 2 r 1 m
2 0f r0 2
− v0 = 1 mv 2
0
2 rf
4 r
0 − 1 Therefore, 14 m 1
2
2
W = (0.90 kg)(22 m/s) 4 − 1 = 5.4 ×10 J 2
16 m ______________________________________________________________________________ 28. REASONING It is useful to divide this problem into two parts. The first part involves the
skier moving on the snow. We can use the workenergy theorem to find her speed when she
comes to the...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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