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Unformatted text preview: nce the weight is W = mg, the mass is
m = W/g. Therefore, the average power is
1
2 P= 2
mvf2 − 1 mv0
Wvf2
2
=
Time
2 g ( Time ) (1) Using the given values for the weight, final speed, and the time, we find that Wvf2
( 9.0 ×103 N ) ( 20.0 m/s ) = 3.3 ×104 W (44 hp)
P=
=
2 g ( Time )
2 ( 9.80 m/s2 ) ( 5.6 s )
2 b. From Equation (1) it follows that Wvf2
(1.4 ×104 N ) ( 20.0 m/s ) = 5.1×104 W (68 hp)
P=
=
2 g ( Time )
2 ( 9.80 m/s 2 ) ( 5.6 s )
2 66. REASONING The work W done is given by Equation 6.1 as W = ( F cos θ ) s , where F is
the magnitude of the force that you exert on the rowing bar, s is the magnitude of the bar’s
displacement, and θ is the angle between the force and the displacement. Solving this
equation for F gives
F= W
( cos θ ) s (6.1) The work is also equal to the average power P multiplied by the time t, or
W = Pt (6.10a) 326 WORK AND ENERGY Substituting Equation (6.10a) into (6.1) gives
F= W
Pt
=
( cos θ ) s ( cos θ ) s SOLUTION The angle θ is 0°, since the direction of the pulling force is the same as the
displacement of the rowing bar. Thus, the magnitude of the force exerted on the handle is ( 82 W )(1.5 s )
Pt
=
= 1.0 × 102 N
( cos θ ) s ( cos 0° )(1.2 m )
______________________________________________________________________________
F= 67. REASONING The average power developed by the cheetah is equal to the work done by
the cheetah divided by the elapsed time (Equation 6.10a). The work, on the other hand, can
be related to the change in the cheetah’s kinetic energy by the workenergy theorem,
Equation 6.3. SOLUTION
a. The average power is
P= W
t (6.10a) where W is the work done by the cheetah. This work is related to the change in the cheetah’s
2
kinetic energy by Equation 6.3, W = 1 mvf2 − 1 mv0 , so the average power is
2
2 W
P=
=
t
= 1 mv 2
f
2
1
2 2 − 1 mv0
2 t
(110 kg )( 27 m / s )2 − 1
2 4.0 s (110 kg )( 0 m / s )2 = 1.0 × 104 W b. The power, in units of horsepower (hp), is 1 hp 4
P = 1.0 × 10 W = 13 hp 745.7 W ______________________________________________________________________________ ( ) Work
Time
(Equation 6.10a). The net work done on the glider is equal to the change in its kinetic
energy, according to the workenergy theorem W = KE f − KE 0 (Equation 6.3). The glider 68. REASONING The power required to accelerate the glider is found from P = accelerates from rest, so its initial kinetic energy is zero. Therefore, the average power
1 mv 2
KE f
mvf2
f
2
required is P =
=
=
. The constant tension in the cable produces a constant
Time
t
2t Chapter 6 Problems 327 acceleration in the glider, permitting us to determine the time from the expression
x = 1 ( v0 + vf ) t (Equation 2.7), where x is the glider’s displacement.
2
SOLUTION Since the glider starts from rest, we have v0 = 0 m/s. Therefore, solving
2x
2x
2x
=
=
.
x = 1 ( v0 + vf ) t (Equation 2.7) for the time t, we obtain t =
2
v0 + vf 0 m/s + vf vf Substituting mv 2
2x
for t in P = f , we find that the average power required of the winch is
vf
2t
mvf3 (184 kg )( 26.0 m/s )
P=
=
=
= 1.68 ×104 W
4 ( 48.0 m ) 2x 4x
2 vf 3 mvf2 69. SSM REASONING AND SOLUTION In the drawings below, the positive direction is to
the right. When the boat is not pulling a skier, the engine must provide a force F1 to
overcome the resistive force of the water FR. Since the boat travels with constant speed,
these forces must be equal in magnitude and opposite in direction.
– FR + F1 = ma = 0 FR or
FR = F1 F1 (1) When the boat is pulling a skier, the engine must provide a force F2 to balance the resistive
force of the water FR and the tension in the tow rope T.
– FR + F2 – T = ma = 0 T F or
F2 = FR + T R F 2 (2) The average power is given by P = F v , according to Equation 6.11 in the text. Since the
boat moves with the same speed in both cases, v1 = v2, and we have
P1 P 2
=
F1 F2 or F2 = F1 Using Equations (1) and (2), this becomes
F1 + T = F1 P2
P1 P2
P1 328 WORK AND ENERGY Solving for T gives
P T = F1 2 − 1
P 1 The force F1 can be determined from P = F1v1 , thereby giving
1 7.50 × 104 W 8.30 × 104 W P1 P 2
2
− 1 =
− 1 = 6.7 × 10 N 7.50 × 104 W v1 P1
12 m/s ______________________________________________________________________________
T= 70. REASONING AND SOLUTION The following drawings show the freebody diagrams for
the car in going both up and down the hill. The force F R is the combined force of air
resistance and friction, and the forces FU and F D are the forces supplied by the engine in
going uphill and downhill respectively.
Going down the hill Going up the hill FN FN FU FR
FD FR
m g sin θ m g cos θ m g sin θ m g cos θ Writing Newton's second law in the direction of motion for the car as it goes uphill, taking
uphill as the positive direction, we have FU − FR − mg sin θ = ma = 0
Solving for FU , we have FU = FR + mg sin θ Similarly, when the car is going downhill, Newton's second law in the direction of motion
gives
FR − FD − mg sinθ = ma = 0
so that
FD = FR − mg sinθ
Since the car needs 47 hp mo...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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