Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
9th Edition
ISBN: 9781305266292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 7, Problem 67CP

(a)

To determine

The value of x i.e. extension of the spring.

(a)

Expert Solution
Check Mark

Answer to Problem 67CP

The value of x is 3.62m4.3023.4m.

Explanation of Solution

When the puck is set into circular motion then force applied by the spring must be equal to centripetal force to keep the system rotating.

Write the expression for force applied by the spring.

    F=kx                                                                                                            (I)

Here, F is the force applied by the spring, k is spring constant and x is compression in the spring.

Write the expression for centripetal acceleration.

    ac=ω2r

Here, ac is centripetal acceleration, ω is angular speed, and r is distance at which object is kept during circular motion.

Write the expression for r.

    r=xus+x

Here, xus is unstressed distance of the spring.

Write the expression for centripetal force.

    Fc=mac                                                                                                       (II)

Here, Fc is centripetal force and m is the mass.

Substitute ω2r for αc in equation (II).

    Fc=mω2r                                                                                                  (III)

Write the expression for ω .

    ω=2πt

Here, t is time period for which the object is in motion.

Substitute 2πt for ω and xus+x for r in equation (III).

    Fc=m(2πt)2xus+x                                                                                   (IV)

Write the expression for conservation of forces.

    F=Fc

Substitute m(2πt)2xus+x for Fc and kx for F in above equation.

    kx=m(2πt)2(xus+x)                                                                                 (V)

Substitute 15.5cm for xus, 1.30s for t, 4.30N/m for k in equation (V).

    (4.30N/m)x=m(2π1.30s)2((15.5cm)(1m100cm)+x)

Simplify above equation.

    (4.30)x=m(23.36)(0.155m+x)3.62m=(4.30)x23.36mx

Rearrange above equation for x .

    x=3.62m4.3023.36m3.62m4.3023.4m                                                                                      (VI)

Conclusion:

Thus, the value of x is (3.62m)m4.30kg(23.4)m.

(b)

To determine

The value of x for mass.

(b)

Expert Solution
Check Mark

Answer to Problem 67CP

The value of x for mass 0.070kg is 0.0951m.

Explanation of Solution

Conclusion:

Substitute 0.070kg for m in equation (VI).

    x=(3.62 m)(0.070kg)4.30 kg23.4(0.070kg)=0.0951m

Thus, the value of x for mass 0.070kg is 0.0951m.

(c)

To determine

The value of x for mass 0.140kg.

(c)

Expert Solution
Check Mark

Answer to Problem 67CP

The value of x for mass 0.140kg is 0.492m .

Explanation of Solution

Conclusion:

Substitute 0.140kg for m in equation (VI).

    x=(3.62 m)(0.140kg)4.30 kg23.36(0.140kg)=0.492m

Thus, the value of x for mass 0.140kg is 0.492m.

(d)

To determine

The value of x for mass 0.180kg.

(d)

Expert Solution
Check Mark

Answer to Problem 67CP

The value of x for mass 0.180kg is 6.85m.

Explanation of Solution

Conclusion:

Substitute 0.180kg for m in equation (VI).

    x=(3.62 m)(0.180kg)4.30 kg23.36(0.180kg)=6.8445m6.85m

Thus, The value of x for mass 0.180kg is 6.85m.

(e)

To determine

The value of x for mass 0.190kg.

(e)

Expert Solution
Check Mark

Answer to Problem 67CP

The value of x for mass 0.190kg is 4.97m which is impossible in this case as the value is negative and this is beyond infinity.

Explanation of Solution

Conclusion:

Substitute 0.190kg for m in equation (VI).

    x=(3.62 m)(0.190kg)4.30 kg23.36(0.190kg)=4.97m

Thus, the value of x for mass 0.190kg is 4.97m which is impossible in this case as the value is negative and this is beyond infinity.

(f)

To determine

Pattern of variation of x for m.

(f)

Expert Solution
Check Mark

Answer to Problem 67CP

The extension of the string x increases with m till m=0.184kg beyond this value, extension is infinity.

Explanation of Solution

The extension of the string x is directly proportional to mass when mass is only in few grams so when mass increases, extension grows faster till m=0.184kg as at this value extension becomes infinity.

When the denominator of the fraction goes to zero, then value of x becomes infinite.

    4.3023.4m=0

Solve above equation for m .

    m=0.184kg

Therefore for m=0.184kg value of extension is infinity so beyond this, situation not possible.

Conclusion:

Thus, the extension of the string x increases with m till m=0.184kg beyond this value , extension is infinity.

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Review.A light spring has unstressed length 15.5 cm.It is described by Hooke's law with spring constant 4.30 N/m. One end of the horizontal spring is held on a fixed vertical axle,and the other end is attched to a puck of mass m that can mow without friction over a horizontal surfaces.The pck is set into motion in a circle with a period pf 1.03 s. (a) Find the extension of the spring x as it depends on m.Evaluate x for (b) m =0.0700kg. (c) m = 0.140 kg  (d) m = 0.180 kg (e) m = 0.190 kg (f)  describe the pattern of variation of x axis as it depnds on m
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Chapter 7 Solutions

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

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