Respuesta :
The initial potential energy of the wagon containing gold boxes will enable
it roll down the hill when cut loose.
The Lone Ranger and Tonto have approximately 5.1 seconds.
Reasons:
Mass wagon and gold = 166 kg
Location of the wagon = 77 meters up the hill
Slope of the hill = 8°
Location of the rangers = 41 meters from the canyon
Mass of Lone Ranger, m₁ = 65 kg
Mass of Tonto m₂ = 66 kg
Solution;
Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m
Potential energy = m·g·h
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912
Potential energy of wagon, P.E. ≈ 17457.0912 J
By energy conservation, P.E. = K.E.
[tex]K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}[/tex]
Where;
v = The velocity of the wagon a the bottom of the cliff
Therefore;
[tex]\dfrac{1}{2} \times 166 \times v^2 = 17457.0912[/tex]
[tex]v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5[/tex]
Velocity of the wagon, v ≈ 14.5 m/s
Momentum = Mass, m × Velocity, v
Initial momentum of wagon = m·v
Final momentum of wagon and ranger = (m + m₁ + m₂)·v'
By conservation of momentum, we have;
m·v = (m + m₁ + m₂)·v'
[tex]\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2) }}[/tex]
Which gives;
[tex]\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66) } \approx 8.1[/tex]
The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s
[tex]Time = \dfrac{Distance}{Velocity}[/tex]
[tex]\mathrm{The \ time \ the\ Lone \ Ranger \ and \ Tonto \ have, \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s[/tex]
The Lone Range and Tonto have approximately 5.1 seconds to grab the
gold and jump out of the wagon before the wagon heads over the cliff.
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