Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 6, Problem 116P

(a)

To determine

The distance moved by person J to reach the lowest point.

(a)

Expert Solution
Check Mark

Answer to Problem 116P

The distance moved by person J to reach the lowest point is h=LLcos20°.

Explanation of Solution

Figure 1 represents the motion of person J.

Physics, Chapter 6, Problem 116P

Here, L is the length of the vine and h is the vertical displacement of person J

Conclusion:

Refer Figure 1.

Write an expression for L.

L=h+Lcos20°

Rewrite equation (I) to find h.

h=LLcos20°

Thus, the distance moved by person J to reach the lowest point is h=LLcos20°.

(b)

To determine

The speed of person J at the lowest point.

(b)

Expert Solution
Check Mark

Answer to Problem 116P

The speed of person J at the lowest point is 4.9m/s.

Explanation of Solution

Write an expression to calculate the conservation of mechanical energy of person J.

               Ki+Ui=Kf+Uf12mvi2+mghi=12mvf2+mghf12vi2+ghi=12vf2+ghf12vi2+g(h)=12vf2+g(0)

12vi2+gh=12vf212vi2+g(LLcos20°)=12vf212vi2+gL(1cos20°)=12vf2 (I)

Here, Ki is the initial kinetic energy of the person J, Ui is the initial potential energy, Kf is the final kinetic energy of the person, Uf is the final potential energy of the person, m is the mass of the person, g is the acceleration due to gravity, hi is the initial height of the person, hf is the final height of the person, vi is the initial velocity of the person and vf is the final velocity of the person.

Rearrange the equation (I) to find vf.

vf=vi2+2gL(1cos20°) (II)

Conclusion:

Substitute 7.0m for L, 4.0m/s for vi and 9.8m/s2 for g in equation (II) to find vf.

vf=(4.0m/s)2+2(9.8m/s2)(7.0m)(1cos20°)=(16m2/s2)+(8.2m2/s2)=4.9m/s

Thus, the speed of person J at the lowest point is 4.9m/s.

(c)

To determine

The height at which person J can swing about the lowest point.

(c)

Expert Solution
Check Mark

Answer to Problem 116P

The height at which person J can swing about the lowest point is 1.24m.

Explanation of Solution

Write an expression to calculate the conservation of mechanical energy of person J.

Ki+Ui=Kf+Uf12mvi2+mghi=12mvf2+mghf12vi2+ghi=12vf2+ghf12vi2+g(h)=12(0)2+ghf12vi2+gh=ghf12vi2+g(LLcos20°)=ghf12vi2+gL(1cos20°)=ghf (III)

Rearrange the equation (III) to find hf.

hf=12gvi2+L(1cos20°) (IV)

Conclusion:

Substitute 7.0m for L, 4.0m/s for vi and 9.8m/s2 for g in equation (IV) to find hf.

hf=12(9.8m/s2)(4.0m/s)2+(7.0m)(1cos20°)=0.81m+0.42m=1.24m

Thus, the height at which person J can swing about the lowest point is 1.24m.

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