Physics

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 130 N at an angle of 30.0° above the horizontal. The box has a mass of 24.0 kg, and the coefficient of kinetic friction between box and floor is 0.300.
(a) Find the acceleration of the box.

(b) The student now starts moving the box up a 10.0° incline, keeping her 130 N force directed at 30.0° above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box?
m/s2, up the incline

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  1. Wb = mg = 24kg * 9.8N/kg = 235 N. = Wt.
    of box.

    Fb = 235N @ 0 Deg. = Force of box.
    Fp = 235*sin(0) = 0 = Force parallel to
    floor.
    Fv = 235*cos(0) = 235 N. = Force perpendicular to floor.

    Fv' = Fv - 130*sin30 = 235 - 65 = 170 N. = Normal.

    Fk = u*Fv' = 0.3*170 = 51 N' = Force of kinetic friction.

    a. Fn = Fap - Fp - Fk = ma.
    130*cos30 - 0 - 51 = 24*a.
    112.58 - 51 = 24a.
    24a = 61.58.
    a = 61.58 / 24 = 2.57 m/s^2.

    b. Fp = 235*sin10 = 40.81 N.
    Fv = 235*cos10 = 231.43 N.
    Fv' = Fv - 130*sin30 = 231.43-65 = 166.43 N.

    Fk = u*Fv' = 0.3*166.43.43 = 49.93 N.

    Fn = Fap - Fp - Fk = ma.
    130*cos30 - 40.81 - 49.93 = 24a.
    112.58 - 40.81 - 49.93 = 24a.
    24a = 112.58 - 90.74 = 21.84.
    a = 21.84 / 24 = 0.91 m/s^2.

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