Unformatted text preview: KE 0 (Equation 6.3). We will use these two
relationships to determine the car’s mass.
SOLUTION Combining KE = 1 mv 2 (Equation 6.2) and W = KE f − KE 0 (Equation 6.3),
2
we obtain
2
1 m v2 − v2 = W
or
W = 1 mvf2 − 1 mv0
f
0
2
2
2 ( ) Solving this expression for the car’s mass m, and noting that 185 kJ = 185 000 J, we find
that
2 (185 000 J )
2W
m= 2 2 =
= 1450 kg
vf − v0 ( 28.0 m/s )2 − ( 23.0 m/s )2 15. SSM WWW REASONING AND SOLUTION The work done on the arrow by the
bow is given by
W = Fs cos 0° = Fs
This work is converted into kinetic energy according to the work energy theorem.
2
W = 1 mvf2 − 1 mv0
2
2 Solving for vf, we find that Chapter 6 Problems 2W
2
+ v0 =
m vf = 2 Fs
2
+ v0 =
m 2 ( 65 N )( 0.90 m ) 291 + ( 0 m/s ) = 39 m/s
2 75 × 10 kg
______________________________________________________________________________
–3 16. REASONING
a. Because we know the flea’s mass m, we can find its speed vf once we know its kinetic
energy KE f = 1 mvf2 (Equation 6.2) at the instant it leaves the ground. The workenergy
2 theorem W = KE f − KE 0 (Equation 6.3) relates KEf to the total work W done on the flea
during pushoff. Because it starts from rest, the flea has no initial kinetic energy, and we see
that the total work done on the flea is equal to its final kinetic energy:
W = KE f − ( 0 J ) = KE f = 1 mvf2 . Since air resistance and the flea’s weight are being
2
ignored, the only force doing work on the flea is the upward force Fground exerted by the
ground, so Wground = W = 1 mvf2 .
2 ( ) b. We will make use of the definition of work Wground = Fground cos θ s (Equation 6.1) to
calculate the flea’s upward displacement s while the force Fground is doing work on the flea.
SOLUTION
a. Solving Wground = 1 mvf2 for the final speed vf of the flea, we find
2 1
2 mvf2 = Wground vf2 = or 2Wground or m vf = 2Wground
m Therefore, vf = ( 2 +2.4 ×10−4 J
1.9 ×10 −4 kg ) = 1.6 m/s b. The force Fground of the ground on the flea is upward, the same as the direction of the
flea’s displacement s while it is pushing off. Therefore, the angle θ between Fground and s is ( ) zero. Solving Wground = Fground cos θ s for the displacement magnitude s, we obtain
s= Wground
Fground cos θ = +2.4 × 10 −4 J ( 0.38 N ) ( cos 0 o ) = 6.3 × 10−4 m 292 WORK AND ENERGY 17. REASONING The work done by the catapult Wcatapult is one contribution to the work done
by the net external force that changes the kinetic energy of the plane. The other contribution
is the work done by the thrust force of the plane’s engines Wthrust. According to the workenergy theorem (Equation 6.3), the work done by the net external force Wcatapult + Wthrust is
equal to the change in the kinetic energy. The change in the kinetic energy is the given
kinetic energy of 4.5 × 107 J at liftoff minus the initial kinetic energy, which is zero since
the plane starts at rest. The work done by the thrust force can be determined from Equation
6.1 [W = (F cos θ) s], since the magnitude F of the thrust is 2.3 × 105 N and the magnitude s
of the displacement is 87 m. We note that the angle θ between the thrust and the
displacement is 0º, because they have the same direction. In summary, we will calculate
Wcatapult from Wcatapult + Wthrust = KEf − KE0.
SOLUTION According to the workenergy theorem, we have
Wcatapult + Wthrust = KEf − KE0
Using Equation 6.1 and noting that KE0 = 0 J, we can write the work energy theorem as
follows:
Wcatapult + ( F cos θ ) s = KE f
14 3
24
Work done by thrust Solving for Wcatapult gives Wcatapult = KE f − ( F cosθ ) s
14 3
24
Work done by thrust = 4.5 ×107 J − ( 2.3 ×105 N ) cos 0° ( 87 m ) = 2.5 ×107 J 18. REASONING AND SOLUTION From the workenergy theorem, Equation 6.3,
1
1
1
2
2
W = mvf2 − mv0 = m vf2 − v0 2
2
2 1
2
2
a. W = (7420 kg) (8450 m/s) – (2820 m/s) =
2 2.35 × 1011 J b. W = (7420 kg) (2820 m/s) − (8450 m/s) = –2.35 × 10 J
2 ______________________________________________________________________________
1 2 2 11 Chapter 6 Problems 19. 293 SSM REASONING The work done to launch either object can be found from
1 1 2 2 Equation 6.3, the workenergy theorem, W = KEf − KE 0 = mvf − mv0 .
2
2
SOLUTION
a. The work required to launch the hammer is ( ) 2
2
W = 1 mvf2 − 1 mv0 = 1 m vf2 − v0 = 1 (7.3 kg) (29 m/s)2 − ( 0 m/s ) = 3.1×103 J
2
2
2
2 2 b. Similarly, the work required to launch the bullet is ( ) 2
W = 1 m vf2 − v0 = 1 (0.0026 kg) (410 m/s) 2 − ( 0 m/s ) = 2.2 × 102 J
2
2 2 ______________________________________________________________________________
20. REASONING
a. The work W done by the force is related to the change in the asteroid’s kinetic energy by
2
the workenergy theorem W = 1 mvf2 − 1 mv0 (Equation 6.3). Since the final speed of the
2
2
asteroid is vf = 5500 m/s and is less than the initial speed of v0 = 7100 m/s, the work
obtained from Equation 6.3 will be negative. This is consistent with the fact that, since the
asteroid is slowing down, it is decelerating, so that the force...
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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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