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Unformatted text preview: each particle; in other words, q0 = ne . Substituting this value for q0 into Equation (2), we
find that t= 5. q0 (VA − VB )
P = ne (VA − VB )
P (3.0 ×1022 ) (1.6 ×10−19 C ) (1.5 V ) = 1800 s
=
4.0 W REASONING The energy to accelerate the car comes from the energy stored in the battery
pack. Work is done by the electric force as the charge moves from point A (the positive
terminal), through the electric motor, to point B (the negative terminal). The work WAB done
by the electric force is given by Equation 19.4 as the product of the charge and the potential
difference VA − VB, or WAB = q0(VA − VB). The power supplied by the battery pack is the
work divided by the time, as expressed by Equation 6.10a.
SOLUTION According to Equation 6.10a, the power P supplied by the battery pack is P= WAB
t = q0 (VA − VB )
t = (1200 C ) ( 290 V ) = 5.0 ×104 W
7.0 s Since 745.7 W = 1 hp (see the page facing the inside of the front cover of the text), the
power rating, in horsepower, is 1 hp
(5.0 ×104 W ) 745.7 W = 67 hp 6. REASONING The electric potential difference ∆V experienced by the electron has the
same magnitude as the electric potential difference experienced by the proton. Moreover,
the charge q0 on either particle has the same magnitude. According to ∆EPE = q0∆V
(Equation 19.4), the losses in EPE for the electron and the proton are the same.
Conservation of energy, then, dictates that the electron gains the same amount of kinetic
energy as does the proton. Both particles start from rest, so the gain in kinetic energy is equal to the final kinetic
energies in each case, which are the same for each particle. However, kinetic energy is
1 mv 2 (Equation 6.2), and the mass of the electron is much less than the mass of the proton
2 1010 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL (see the inside of the front cover). Since the kinetic energies are the same, the speed ve of the
electron will be greater than the speed vp of the proton.
SOLUTION Let us assume that the proton accelerates from point A to point B. According
to the energyconservation principle (including only kinetic and electric potential energies),
we have KE p, B + EPE p, B = KE p, A + EPE p, A
1442443 1442444
4
3
Final total energy
of proton at point B or KE p, B = EPE p, A − EPE p, B Initial total energy
of proton at point A Here we have used the fact that the initial kinetic energy is zero since the proton starts from
rest. According to Equation 19.3, the electric potential energy of a charge q0 is EPE = q0V,
where V is the potential experienced by the charge. We can, therefore, write the final kinetic
energy of the proton as follows:
KE p, B = EPE p, A − EPE p, B = qp (VA − VB ) (1) The electron has a negative charge, so it accelerates in the direction opposite to that of the
proton, or from point B to point A. Energy conservation applied to the electron gives
KEe, A + EPEe, A = KEe, B + EPEe, B
1442443 1442443
Final total energy
of electron at point A or KE e, A = EPEe, B − EPEe, A Initial total energy
of electron at point B Here we have used the fact that the initial kinetic energy is zero since the electron starts
from rest. Again using Equation 19.3, we can write the final kinetic energy of the electron as
follows:
KE e, A = EPE e, B − EPE e, A = qe (VB − VA ) But an electron has a negative charge that is equal in magnitude to the charge on a proton,
so qe = –qp. With this substitution, we can write the kinetic energy of the electron as
KE e, A = EPE e, B − EPE e, A = −qp (VB − VA ) = qp (VA − VB ) (2) Comparing Equations (1) and (2), we can see that
KE p, B = KEe, A or 1 m v2
2 pp 2
= 1 meve
2 Solving for the ratio ve/vp and referring to the inside of the front cover for the masses of the
electron and the proton, we obtain Chapter 19 Problems ve
vp 7. mp = me = 1011 1.67 ×10−27 kg
= 42.8
9.11× 10−31 kg SSM REASONING The translational speed of the particle is related to the particle’s
translational kinetic energy, which forms one part of the total mechanical energy that the
particle has. The total mechanical energy is conserved, because only the gravitational force
and an electrostatic force, both of which are conservative forces, act on the particle (see
Section 6.5). Thus, we will determine the speed at point A by utilizing the principle of
conservation of mechanical energy. SOLUTION The particle’s total mechanical energy E is
1
mv 2
2
1 24
43 E= + Translational
kinetic
energy 1
Iω 2 +
2
1 24
43
Rotational
kinetic
energy 1 mgh + kx 2 + EPE
2
123 1 24 123
4 3 44
Gravitational
potential
energy Elastic
potential
energy Electric
potential
energy Since the particle does not rotate, the angular speed ω is always zero and since there is no
elastic force, we may omit the terms 1
2 I ω 2 and 1
2 kx 2 from this expression. With this in mind, we express the fact that EB = E A (energy is conserved) as follows:
1
2
mvB
2 1
2 2
+ mghB + EPE B = mv A + mghA + EPE A This equation can be simplified further, since th...
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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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