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Korvikt [17]
3 years ago
12

A 77.0 kg woman slides down a

Physics
1 answer:
statuscvo [17]3 years ago
3 0

Assuming the woman starts at rest, she descends the slide with acceleration <em>a</em> such that

(20.3 m/s)² = 2 <em>a</em> (42.6 m)   →   <em>a</em> ≈ 4.84 m/s²

which points parallel to the slide.

The only forces acting on her, parallel to the slide, are

• the parallel component of her weight, <em>w</em> (//)

• friction, <em>f</em>, opposing her descent and pointing up the slide

Take the downward sliding direction to be positive. By Newton's second law, the net force in the parallel direction acting on the woman is

∑ <em>F</em> (//) = <em>w </em>(//) - <em>f</em> = <em>ma</em>

where <em>m</em> = 77.0 kg is the woman's mass.

Solve for <em>f</em> :

<em>mg</em> sin(42.3°) - <em>f</em> = <em>ma</em>

<em>f</em> = <em>m</em> (<em>g</em> sin(42.3°) - <em>a</em>)

<em>f</em> = (77.0 kg) ((9.80 m/s²) sin(42.3°) - 4.84 m/s²) ≈ 135 N

Compute the work <em>W</em> done by friction: multiply the magnitude of the friction by the length of the slide.

<em>W</em> = (135 N) (42.6 m) ≈ 5770 N•m = 5770 J

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Answer:

<em>0.85c </em>

Explanation:

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E = γE_{0}

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E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

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