Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 17, Problem 49P
To determine

Arrange the change in kinetic energy in the decreasing order.

Expert Solution & Answer
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Answer to Problem 49P

The decreasing order of change in kinetic energy is (c), (b), (e), (d), (a)=(f).

Explanation of Solution

Write the expression for conservation of energy

Kf+Uf=Ki+Ui                                                              (I)

Here, Kf is the final kinetic energy, Ki is the initial kinetic energy, Uf is the final potential energy and Ui is the initial kinetic energy.

Rearrange (I)

KfKi=UiUf

Write the expression for change in kinetic energy

ΔK=ΔU                                                  (II)

Here, ΔK is the change in kinetic energy and ΔU is the change in electric potential energy in moving the charge from P to S.

Write the expression for ΔU

ΔU=q(VfVi)                                                  (III)

Here, q is the charge being moved, Vf is the potential at S  and Vi is the potential at the point, P.

Substitute (III) in (II)

ΔK=q(VfVi)                                                  (IV)

(a)Substitute  5nC for q, 100V for Vi and 50V for Vf in (IV) to find ΔK

ΔK=(5nC)(50V(100V))=(5×109C)(50V(100V))=750×109 J

Thus, the change in kinetic energy in (a) is 750×109 J.

(b)Substitute  5nC for q, 50V for Vi and 50V for Vf in (IV) to find ΔK

ΔK=(5nC)(50V(50V))=(5×109C)(50V(50V))=500×109 J

Thus, the change in kinetic energy in (b) is 500×109 J.

(c)Substitute  25nC for q, 50V for Vi and 20V for Vf in (IV) to find ΔK

ΔK=(25nC)(20V(50V))=(25×109C)(20V(50V))=750×109 J

Thus, the change in kinetic energy in (c) is 750×109 J.

(d)Substitute  1nC for q, 400V for Vi and 100V for Vf in (IV) to find ΔK

ΔK=(1nC)(100V(400V))=(1×109C)(100V(400V))=500×109 J

Thus, the change in kinetic energy in (d) is 500×109 J.

(e)Substitute  1nC for q, 100V for Vi and 250V for Vf in (IV) to find ΔK

ΔK=(1nC)(250V(100V))=(1×109C)(250V(100V))=150×109 J

Thus, the change in kinetic energy in (e) is 150×109 J.

(e)Substitute  5nC for q, 100V for Vi and 250V for Vf in (IV) to find ΔK

ΔK=(5nC)(250V(100V))=(5×109C)(250V(100V))=750×109 J

Thus, the change in kinetic energy in (f) is 750×109 J.

Thus, the decreasing order of change in kinetic energy is (c), (b), (e), (d), (a)=(f).

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Chapter 17 Solutions

Physics

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Charges of + 2.0 nC and − 1.0 nC are located...Ch. 17 - Prob. 27PCh. 17 - 28. (a) Find the potential at points a and b in...Ch. 17 - 29. (a) In the diagram, what are the potentials at...Ch. 17 - 30. (a) In the diagram, what are the potentials at...Ch. 17 - Prob. 31PCh. 17 - 32. By rewriting each unit in terms of kilograms,...Ch. 17 - 33. Rank points A–E in order of the potential,...Ch. 17 - Prob. 34PCh. 17 - Prob. 35PCh. 17 - Prob. 36PCh. 17 - Prob. 37PCh. 17 - Prob. 38PCh. 17 - Prob. 39PCh. 17 - Prob. 40PCh. 17 - Prob. 41PCh. 17 - Prob. 43PCh. 17 - 43. A positive point charge is located at the...Ch. 17 - Prob. 44PCh. 17 - Prob. 45PCh. 17 - 46. Point P is at a potential of 500.0 kV, and...Ch. 17 - 47. An electron is accelerated from rest through a...Ch. 17 - 48. As an electron moves through a region of...Ch. 17 - Prob. 49PCh. 17 - 50. An electron beam is deflected upward through...Ch. 17 - 51. In the electron gun of Example 17.8, if the...Ch. 17 - 52. 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