A mass of 4 kg is traveling over a quarter circular ramp with a radius of 10 meters. At the bottom of the incline the mass is moving at 21.

Question

A mass of 4 kg is traveling over a quarter circular ramp with a radius of 10 meters. At the bottom of the incline the mass is moving at 21.3 m/s and at the top of the incline the mass is moving at 2.8 m/s. What is the work done by all non-conservative force in Joules?

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Nem 7 months 2021-07-19T04:21:44+00:00 1 Answers 4 views 0

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    2021-07-19T04:23:10+00:00

    Answer:

    499.7 J

    Explanation:

    Since total mechanical energy is conserved,

    U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.

    So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂

    mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

    Substituting the values of the variables into the equation, we have

    mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

    4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)²  + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)²  + W₂

    0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s²  + W₂

    907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s²  + W₂

    907.38 kgm²/s² = 407.68 kgm²/s² + W₂

    W₂ = 907.38 kgm²/s² – 407.68 kgm²/s²

    W₂ = 499.7 kgm²/s²

    W₂ = 499.7 J

    Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J

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