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26 July, 05:50

Use the work-energy theorem to solve each of these problems. You can use Newton's laws to check your answers. (a) A skier moving at 5.00 m/s encounters a long, rough horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping?

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  1. 26 July, 07:09
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    d = 5.8 m

    Explanation:

    Principle of work and energy

    ΔE = Wf

    ΔE = Ef-Ei

    ΔE : mechanical energy change

    Wf : Work done by kinetic friction force

    Ef : final mechanical energy

    Ei : initial mechanical energy

    K = (1/2) mv² : Kinetic energy

    U = mgh : Potential energy

    m: mass (kg)

    v : speed (m/s)

    h: high (m)

    Data

    vi = 5.00 m/s

    vf=0

    hi=0

    hf=0

    μk=0.220

    Kinetic friction force

    fk = μk * FN

    FN = W : normal force (N)

    W = m*g : weight (N)

    FN = 9.8*m (N)

    fk = 0.220*9.8*m

    fk = (2.156) * m Equation (1)

    Work done by the kinetic friction force

    Wf = - fk*d (J) Equation (1)

    d: distance traveled by force

    Principle of work and energy to the skier

    ΔE = Ef-Ei

    Ef = Kf + Uf = 0

    Ei = Kf + Uf = (1/2) (m) (5) ² + 0

    ΔE = - (1/2) (m) (5) ²

    ΔE = Wf

    0 - (1/2) (m) (5) ² = - fk*d

    We replace fk = (2.156) * (m) of the equation (1)

    - (1/2) (m) (5) ² = - (2.156) * (m) * d

    We divide by (-m) both sides of the equation

    (1/2) (5) ² = (2.156) * d

    12.5 = (2.156) * d

    d = (12.5) / (2.156)

    d = 5.8 m
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