A 116 kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.85 m higher than the surface o

Question

A 116 kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.85 m higher than the surface of the water and the ramp is inclined at an angle of 26.5 ∘ above the horizontal.

Part A If the seal reaches the water with a speed of 4.10 m/s, what is the work done by kinetic friction?

Part B What is the coefficient of kinetic friction between the seal and the ramp?

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Thu Thủy 2 months 2021-07-29T20:47:09+00:00 2 Answers 4 views 0

Answers ( )

    0
    2021-07-29T20:48:42+00:00

    Answer:

    a) W_{fr} = 1129.602\,J, b) \mu_{k} = 0.268

    Explanation:

    The seal is modelled after the Principle of Energy Conservation and the Work-Energy Theorem:

    U_{g,A} + K_{A} = U_{g,B} + K_{B} + W_{fr}

    a) The work done by the kinetic friction is:

    W_{fr} = U_{g,A}-U_{g,B}+K_{A}-K_{B}

    W_{fr} = (116\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.85\,m-0\,m) + \frac{1}{2}\cdot (116\,kg)\cdot \left[(0\,\frac{m}{s} )^{2} - (4.10\,\frac{m}{s} )^{2}\right]

    W_{fr} = 1129.602\,J

    b) The coefficient of kinetic friction is:

    W_{fr} = \frac{\mu_{k}\cdot (m\cdot g \cdot \cos \theta)\cdot (h_{A}-h_{B})}{\sin \theta}

    \mu_{k} = \frac{W_{fr}\cdot \sin \theta}{(m\cdot g \cdot \cos \theta)\cdot (h_{A}-h_{B})}

    \mu_{k} = \frac{1129.602\,J\cdot \sin 26.5^{\textdegree}}{(116\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (\cos 26.5^{\textdegree})\cdot (1.85\,m)}

    \mu_{k} = 0.268

    0
    2021-07-29T20:48:54+00:00

    Answer:

    (A) Work done by friction, Wf = 1128.1 J

    (B) The coefficient of kinetic friction between the seal and the ramp,μ is 0.268

    Explanation:

    Given;

    mass of seal, m = 116 kg

    top of the ramp, h = 1.85 m

    angle of inclination, θ = 26.5⁰

    Part A

    Apply the principle of conservation of Energy;

    sum of initial potential energy and initial kinetic energy = sum of final potential energy, final kinetic energy and work lost due to friction.

    U_i + K_i = U_f + K_f + W_{friction}

    Initial kinetic energy is zero, since the seal slides from rest.

    Final potential energy is zero, at the end of the ramp, height is zero.

    The equation above reduces to;

    U_i = k_f + W_{friction}

    Mgh = ¹/₂Mv² + Wf

    Wf = Mgh – ¹/₂Mv²

    Wf = (116 x 9.8 x 1.85) – (¹/₂ x 116 x 4.1²)

    Wf = 2103.08 – 974.98 = 1128.1 J

    Thus, work done by friction, Wf = 1128.1 J

    Part B

    work done by friction, Wf = Frictional force x distance moved by the seal down the slope

    Wf  = Fk x d

    Wf = μmgcosθ x d

    d is the slope of the inclined ramp to the horizontal, this calculated using trigonometry ratio;

    d (slope) = h/sinθ

    d = 1.85/sin26.5

    d = 4.146 m

    Wf = μmgcosθ x d

    1128.1 = μ x 116 x 9.8 x cos(26.5) x 4.146

    1128.1  = μ x 4217.98

    μ = 1128.1 / 4217.98

    μ = 0.268

    Thus, the coefficient of kinetic friction between the seal and the ramp is 0.268

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