Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

help with physics?

An Eskimo pulls a 150.0kg sled with a constant force through the snow at a constant speed of 2.0m/s.

He applies this force by pulling on a rope at 30◦

to the horizontal. There is a coefficient of kinetic friction between

the sled and the snow of µk = 0.1.

(a) Find the tension in the rope

(b) What is the work done by the Eskimo as he moves the sled 100m across the ground?

(c) If the Eskimo suddenly stops pulling, but the coefficient of kinetic friction remains the same, how far

does the box travel before it comes to rest (assume the Eskimo gets out of the way)?

(d) What is the power due to the frictional force (rate at which energy is being dissipated) the instant he

stops pulling on the box?

1 Answer

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  • Whome
    Lv 7
    1 month ago
    Favourite answer

    (a) Find the tension in the rope

    Normal force N of snow in sled

    N = mg - Tsinθ

    Friction force Ff of snow on sled

    Ff = μΝ = μmg - μTsinθ

    As velocity is constant, the friction force must equal the horizontal component of tension

                    Tcosθ = μmg - μTsinθ 

     Tcosθ + μTsinθ = μmg

    T(cosθ + μsinθ) = μmg

                           T = μmg/(cosθ + μsinθ)

                           T = 0.1(150)(9.8) / (cos30 + 0.1sin30)

                           T = 160.47589... ≈ 160 N

    (b) What is the work done by the Eskimo as he moves the sled 100m across the ground?

    W = Fd = (Tcosθ)d = (160cos30)100 = 13,897.620... ≈ 14 kJ

    (c) If the Eskimo suddenly stops pulling, but the coefficient of kinetic friction remains the same, how far does the box travel before it comes to rest? 

    The kinetic energy will convert to work of friction.

         Fd = ½mv²

    μmgd = ½mv²

           d = v²/2μg = 2.0²/(2(0.1)(9.8)) = 2.0408...  ≈ 2.0 m

    (d) What is the power due to the frictional force?

    Power is the rate of doing work.

    The work done equals the initial kinetic energy. = ½(150)2.0² = 300 J

    The time needed to stop is v/a = v/(Ff/m) = v/(μmg/m) = v/μg = 2/((0.1)9.8) = 2.0408...  ≈ 2.0 s

    P = 300 J / 2.0 s = 150 W(atts)

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