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mr Goodwill [35]
3 years ago
7

A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is

the maximum vertical height the person in the cart can reach?
Physics
1 answer:
HACTEHA [7]3 years ago
5 0

Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m

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Answer:

μsmín = 0.1

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       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

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     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

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       m* g = m* \mu_{smin} * \omega^{2} * r (4)

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       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

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