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23 December, 06:03

A 42.0 kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.95 m higher than the surface of the water and the ramp is inclined at an angle of 35.0 degrees above the horizontal. A) If the seal reaches the water with a speed of 4.30 m/s, what is the work done by kinetic friction? B) What is the coefficient of kinetic friction between the seal and the ramp?

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  1. 23 December, 10:02
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    A) Wf = - 414.33 J

    B) μk=0.36

    Explanation:

    Newton's second law

    ∑F = m*a Formula (1)

    ∑F : algebraic sum of the forces in Newton (N)

    m : mass in kilograms (kg)

    a : acceleration in meters over second square (m/s²)

    Known data

    m=42.0 kg mass of the seal

    h = 1.95 m : hight of the ramp

    θ = 35° angle of the ramp with respect to the horizontal

    g = 9.8 m/s² : acceleration due to gravity

    Forces acting on the seal

    We define the x-axis in the direction parallel to the movement of the seal on the ramp and the y-axis in the direction perpendicular to it.

    W: Weight of the seal : In vertical direction downward

    FN : Normal force : perpendicular to the direction the ramp

    fk : Friction force: parallel to the direction to the ramp

    Calculated of the weight of the seal

    W = m*g = (42 kg) * (9.8 m/s²) = 411.6 N

    x-y weight components

    Wx = Wsin θ = (411.6) * sin (35) °=236.08 N

    Wy = Wcos θ = (411.6) * cos (35) ° = 337.16 N

    Calculated of the Normal force

    ∑Fy = m*ay ay = 0

    FN-Wy = 0

    FN=Wy = 337.16 N

    Calculated of the Friction force:

    fk=μk*FN = μk * 337.16 N Equation (1)

    A) Principle of work and energy

    ΔE = Wf

    ΔE:mechanical energy change

    Wf: Work done by kinetic friction force

    K : Kinetic energy

    U: Potential energy

    ΔE = Wf

    Ef-Ei = Wf

    (K+U) final - (K+U) initial = Wf

    ((1/2) mv²+0) - (0+m*g*h) = Wf

    (1/2) (42) (4.3) ² - (42) (9.8) (1.95) = Wf

    388.29-802.62 = Wf

    Wf = - 414.33 J

    B) The coefficient of kinetic friction between the seal and the ramp

    Wf = f*d Equation (2)

    d: length of the ramp

    sin θ = h/d

    d = h / sin θ = 1.95 / sin 35

    d = 3.4 m

    We replace Wf = - 414.33 J and d = 3.4 m in the equation (2)

    -414.33 J = - f * (3.4m)

    fk = 414.33 N*m / 3.4m

    fk=121.86 N

    Calculated of the coefficient kinetic friction

    μk=fK/FN

    μk=121.86 N/337.16 N

    μk=0.36
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