Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 8, Problem 61AP

(a)

To determine

The total energy of the child stick Earth system .

(a)

Expert Solution
Check Mark

Answer to Problem 61AP

The total energy of the child-stick Earth system is 100J.

Explanation of Solution

Refer to Figure P8.61; Total energy of the system remains conserved at every point in the system.

Consider point A to calculate total energy. As the boy is momentarily at rest at point A Kinetic energy for the system is zero.

Write the expression for elastic potential energy stored in the spring.

  Us=12kx2                                                                                                     (I)

Here, Us is elastic potential energy, k is force constant and x is compression in the spring.

Write the expression for potential energy.

    Ug=mgh                                                                                                     (II)

Here, Ug is gravitational potential energy, m is mass, g is acceleration due to gravity and h is height.

Write the expression for kinetic energy of the system.

    K=12mv2                                                                                                  (III)

Here, K is kinetic energy and v is velocity.

Write the expression for total energy at point A.

    Etotal=KA+UgA+UsA                                                                               (IV)

Here, Etotal is the total energy of the system, KA is kinetic energy at point A, UgA is gravitational potential energy at point A, UsA is the elastic potential energy at point A.

Substitute 12mvA2 for KA , 12kxA2 for UsA and mgxA for UgA in equation (IV)

    Etotal=12mvA2+mgxA+12kxA2

Here, vA is speed at A and xA is distance moved by the system at point A which is equal to compression of the string.

Substitute  0 for 12mvA2 in above equation.   

    Etotal=0+mgxA+12kxA2                                                                            (V)

Conclusion:

Substitute  2.50×104N/m for k  , 0.100m for xA , 9.80m/s2 for g and 25.0kg for m in equation (V).

    Etotal=[0+(25.0kg)(9.80m/s2)(0.100m)+12(2.50×104N/m)(0.100m)2]=24.5+125=100.5J100J

Thus, the total energy of the child-stick Earth system is 100J.

(b)

To determine

The value of xc.

(b)

Expert Solution
Check Mark

Answer to Problem 61AP

The value of xc is 0.410m .

Explanation of Solution

Total energy at point A is equal to total energy at point C as the energy of system is conserved.

Write the expression for total energy at point C.

    EC=KC+UgC+UsC                                                                                   (VI)

Here, EC is the energy of the system at C, KC is kinetic energy at point C, UgC is gravitational potential energy at point C, and UsC is the elastic potential energy at point C.

Kinetic energy at C is zero as child again momentarily comes to rest at the top of the jump and elastic potential energy at C is also zero because string is no more compressed.

Substitute 12mvC2 for KC , 12kx2 for UsC and mgxC for UgC in equation (VI).

    EC=12mvC2+mgxC+12kx2

Here, vC is speed at C and xC is distance moved by the system at point C.

Substitute  0 for 12mvC2 and 0 for 12kx2 in above equation.   

    EC=0+mgxC+0                                                                                      (VII)

Write the expression for conservation of energy.

    Etotal=EC

Substitute mgxC for EC in above equation.

    Etotal=mgxC

Rearrange above equation for xc .

    xC=Etotalmg                                                                                                (VIII)

Conclusion:

Substitute 100.5J for Etotal, 9.80m/s2 for g and 25.0kg for m in equation (VIII).

    xC=(100.5J)(25.0kg)(9.80m/s2)=0.410m

Thus, the value of xc is 0.410m .

(c)

To determine

The speed of the child at point B.

(c)

Expert Solution
Check Mark

Answer to Problem 61AP

Thus, the speed of the child at point B is 2.84m/s.

Explanation of Solution

When x=0  i.e. at point B elastic potential and gravitational potential energy is zero.

Write the expression for total energy at point C.

    EB=KB+UgB+UsB                                                                                   (IX)

Here, EB is the energy of the system at B, KB is kinetic energy at point B, UgB is gravitational potential energy at point B, and UsB is the elastic potential energy at point B.

Substitute 12mvB2 for KB , 12kx2 for UsB and mgxB for UgB in equation (IX).

    EB=12mvB2+mgxB+12kx2

Here, vB is speed at C and xB is distance moved by the system at point B.

Substitute  0 for mgxB and 0 for 12kx2 in above equation.   

    EB=12mvB2+0+0                                                                                       (X)

Write the expression for conservation of energy.

    Etotal=EB

Substitute 12mvB2 for EB in above equation.

    Etotal=12mvB2

Rearrange above equation for vB .

    vB=2Etotalm                                                                                               (XI)

Conclusion:

Substitute 100.5J for Etotal  and 25.0kg for m in equation (XI).

    vB=2(100.5J)(25.0kg)=2.835m/s2.84m/s

Thus, the speed of the child at point B is 2.84m/s.

(d)

To determine

The value of x for which the kinetic energy of the system is a maximum.

(d)

Expert Solution
Check Mark

Answer to Problem 61AP

The value of x for which the kinetic energy of the system is a maximum is 9.80mm.

Explanation of Solution

To calculate the value of x for which the kinetic energy of the system is maximum..

Write the expression for total energy at any point.

    E=K+Ug+Us

Here, E is the total energy of the system .

Rearrange above equation for K .

    K=EUgUs

Substitute 12kx2 for Us and mgx for Ug in above equation.

    K=Emgx12kx2

Substitute  (x) for x in above equation as     x is the compression of the spring its movement is in opposite direction.

    K=Emg(x)12k(x)2

Simplify above expression.

  K=E+mgx12k(x)2                                                                              (XII)   

Differentiate the equation with respect to x and equate the result to zero

Differentiate equation (XII) with respect to x .

    dKdx=0+mgkx

Equate it to zero.

    0+mgkx=0

Rearrange above equation for x .

    x=mgk                                                                                                     (XIII)

Conclusion:

Substitute 2.50×104N/m for k , 9.80m/s2 for g and 25.0kg for m in equation (XIII).

    x=(25.0kg)(9.80m/s2)(2.50×104N/m)=0.0098m(1000mm1m)=9.80mm

As x is value of compression and it is below x=0 value of x is negative. Therefore value of x is 9.80mm.

Thus, the value of x for which the kinetic energy of the system is a maximum is 9.80mm.

(e)

To determine

The child’s maximum upward speed.

(e)

Expert Solution
Check Mark

Answer to Problem 61AP

The child’s maximum upward speed is 2.85m/s.

Explanation of Solution

Consider point D where velocity is maximum and it is the point where we have maximum kinetic energy.

Write the expression for total energy at point D.

    ED=KD+UgD+UsD                                                                               (XIV)

Here, ED is the energy of the system at D, KD is kinetic energy at point D, UgD is gravitational potential energy at point D, and UsD is the elastic potential energy at point D.

Write the expression for conservation of energy between point A and D.

    Etotal=ED

Substitute KD+UgD+UsD for ED in above equation.

    Etotal=KD+UgD+UsD

Rearrange above equation for KD .

  KD=EtotalUgDUsD                                                                           (XV)

Substitute 12mvD2 for KD , 12kxD2 for UsD and mgxD for UgD in equation (XV)

    12mvD2=EtotalmgxD12kxD2

Here, vD is speed at D and xD is distance moved by the system at point D.

Multiply above equation by 2 .

    mvD2=2Etotal2mgxDkxD2

Rearrange above equation for vD

    vD=2Etotal2mgxDkxD2m                                                                 (XVI)

Conclusion:

Substitute 2.50×104N/m for k, 9.80mm for xD , 100.5J for Etotal, 9.80m/s2 for g and 25.0kg for m in equation (XVI).

    vD=[2(100.5J)2(25.0kg)(9.80m/s2)(9.80mm)(1m1000mm)(2.50×104N/m)((9.80mm)(1m1000mm))2](25.0kg)=201J+4.802J -2.401J(25.0kg)=2.85m/s

Thus, the child’s maximum upward speed is 2.85m/s.

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Chapter 8 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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