Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
5th Edition
ISBN: 9781305586871
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 7, Problem 83P

(a)

To determine

Maximum speed of the needle.

(a)

Expert Solution
Check Mark

Answer to Problem 83P

Maximum speed is 21m/s_.

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 7, Problem 83P , additional homework tip  1

Consider the needle is fired horizontally from a spring. Maximum speed occurs just after the needle leaves the spring, before entering the body.

Write the law of conservation of energy for this case.

  Ki+UiE=Kf+Uf        (I)

Here Ki is the initial kinetic energy, Ui is the initial elastic potential energy, Kf is the final kinetic energy Uf is the final elastic potential energy and E is the work done by the spring.

Write the equation for initial and final kinetic energy.

  Ki=12mvi2Kf=12mvf2        (II)

Here m is the mass of the needle, vi is the initial velocity and vf is the final velocity.

Write down the equation for initial and elastic potential energy.

  Ui=12kxi2Uf=12kxf2        (III)

Here k is the spring constant, xi is the initial extension of the spring and xf is the final extension of the spring.

Write the equation for work done by the spring

  E=fkd        (IV)

Here E is the work done by the spring, fk is the force constant of the spring and d is the distance moved into the tissue.

As the initial velocity s zero, Ki=0

As there is no penetration into the body, distance covered is zero. Then E=0

As there is no extension, Uf=0

Substitute these results in (I) and write for vf.

  0+12kxi20=12mvf2+0vf2=kxi2mvf=kmxi        (V)

Conclusion:

Substitute 375N/m for k, 5.6g for m and 8.1cm for xi in (V)

  vf=375N/m5.6g(103kg1g)(8.1cm(102m1cm))=375N/m5.6×103kg(8.1×102m)=20.96m/s21m/s

Maximum speed is 21m/s_.

(b)

To determine

Speed to limit the penetration to 5.9cm.

(b)

Expert Solution
Check Mark

Answer to Problem 83P

Speed should be 16m/s_.

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 7, Problem 83P , additional homework tip  2

The initial elastic potential energy is converted partially into internal energy in the organ and partially kinetic energy of the needle.

Write the energy conservation equation for this case.

  Ki+UiEiEf=Kf+Uf        (VI)

Here Ki is the initial kinetic energy, Ui is the initial elastic potential energy, Kf is the final kinetic energy Uf is the final elastic potential energy Ei is the initial work done by the spring and Ef is the final work done by the spring.

Write the equation for initial and final work done by the spring

  Ei=fkidiEf=fkfdf        (VII)

Here, fki is the initial force constant of the spring, fkf is the final force constant of the spring, di is the initial distance moved into the tissue and df is the final distance moved into the tissue.

As the initial velocity s zero

  Ki=0        (VIII)

As there is no extension,

  Uf=0        (IX)

Substitute (II), (III), (VII), (VIII) and (IX) in (VI) and write for vf.

  12mvf2+0=0+12kxi2fkidifkfdfvf2=2(12kxi2fkidifkfdf)mvf=2(12kxi2fkdifkdf)m        (X)

Conclusion:

Substitute 375N/m for k, 5.6g for m, 8.1cm for xi, 7.6N for fki, 2.4cm for di, 9.2N for fkf and 3.5cm for xf in (X)

  vf=2(12375N/m(8.1cm(102m1cm))2((7.6N)(2.4cm(102m1cm)))((9.2N)(3.5cm(102m1cm))))5.6g(103kg1g)=2(12375N/m(8.1×102m)2((7.6N)(2.4×102m))((9.2N)(3.5×102m)))5.6×103kg=16m/s

Speed should be 16m/s_.

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Chapter 7 Solutions

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)

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