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 6, Problem 65P

(a)

To determine

The extension of the spring for a mass of m.

(a)

Expert Solution
Check Mark

Answer to Problem 65P

The extension of the spring for a mass of m is x=4π2mL0/T2k4π2mL0/T2_.

Explanation of Solution

Write the expression for centripetal force of the mass m attached at the end of the spring.

    F=mv2r        (I)

Here, F is the centripetal force, m is the mass, v is the velocity, r is the radius of the orbit.

Write the expression for velocity in terms of time period.

    v=2πrT        (II)

Here, v is the velocity, r is the radius of the orbit, and T is the time period.

Write the expression for force from hooks law.

    F=kx        (III)

Here, F is the force, k is the spring constant, and x is the distance.

Use equation (II) and (III) in equation (I) and rearrange.

    kx=m(2πrT)2rkT2x=4π2mr        (IV)

Write the expression for radius of the pluck’s motion.

    r=L0+x        (V)

Use equation (V) in equation (IV), to find x.

    kT2x=4π2m(L0+x)kx=(4π2mL0)T2+x(4π2m)T2x(k4π2mT2)=4π2mL0T2x=4π2mL0/T2k4π2mL0/T2        (VI)

Conclusion:

Therefore, the extension of the spring for a mass of m is x=4π2mL0/T2k4π2mL0/T2_.

(b)

To determine

The extension of the spring for the mass 0.0700kg.

(b)

Expert Solution
Check Mark

Answer to Problem 65P

The extension of the spring for the mass 0.0700kg is 0.0951m_.

Explanation of Solution

Substitute 4.30N/m for k, 0.155m for L0, 1.30s for T, in equation (VI), to find x.

    x=4(3.14)2(0.155m)m/(1.30s)24.30N/m4(3.14)2m(0.155m)/(1.30s)2=(3.62m/s2)m4.30kg/s2(23.361/s2)m=(3.62m)m[4.30kg(23.36)]m1/s21/s2=(3.62m)m4.30kg(23.4)m        (VII)

Conclusion:

Substitute 0.070kg for m in equation (VII), to find x.

    x=(3.62m)(0.070kg)4.30kg(23.4)(0.070kg)=0.0951m

Therefore, the extension of the spring for the mass 0.0700kg is 0.0951m_.

(c)

To determine

The extension of the spring for the mass 0.140kg.

(c)

Expert Solution
Check Mark

Answer to Problem 65P

The extension of the spring for the mass 0.140kg is 0.492m_.

Explanation of Solution

From equation (VII).

    x=(3.62m)m4.30kg(23.4)m

Conclusion:

Substitute 0.140kg for m in equation (VII), to find x.

    x=(3.62m)(0.140kg)4.30kg(23.36)(0.140kg)=0.492m

Therefore, the extension of the spring for the mass 0.140kg is 0.492m_.

(d)

To determine

The extension of the spring for the mass 0.180kg.

(d)

Expert Solution
Check Mark

Answer to Problem 65P

The extension of the spring for the mass 0.180kg is 6.85m_.

Explanation of Solution

From equation (VII).

    x=(3.62m)m4.30kg(23.4)m

Conclusion:

Substitute 0.180kg for m in equation (VII), to find x.

    x=(3.62m)(0.180kg)4.30kg(23.36)(0.180kg)=0.6520.0952m=6.85m

Therefore, the extension of the spring for the mass 0.180kg is 6.85m_.

(e)

To determine

The extension of the spring for the mass 0.190kg.

(e)

Expert Solution
Check Mark

Answer to Problem 65P

For the mass 0.190kg the spring cannot constrain the motion, this situation is impossible.

Explanation of Solution

From equation (VII) the spring extension is given by

    x=(3.62m)m4.30kg(23.4)m

Conclusion:

Substitute 0.190kg for m in equation (VII), to find x.

    x=(3.62m)(0.190kg)4.30kg(23.36)(0.190kg)=0.6520m=

Therefore, For the mass 0.190kg the extension of the spring goes to infinity. The spring cannot constrain the motion, this situation is impossible

(f)

To determine

To explain the pattern of variation of x as it depends on m.

(f)

Expert Solution
Check Mark

Answer to Problem 65P

The extension of the spring is directly proportional to the mass m for the mass in few grams. For the mass 0.184kg and more the extension starts to diverge to infinity.

Explanation of Solution

The extension of the spring is directly proportional to the mass m, for the lighter mass in few grams. When the mass is further more increased from 0.184kg the extension starts to diverge to infinity.

Conclusion:

Therefore, the extension of the spring is directly proportional to the mass m for the mass in few grams. For the mass 0.184kg and more the extension starts to diverge to infinity.

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Chapter 6 Solutions

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

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