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Physics

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A 13.0 kg box is released on a 33 degree incline and accelerates down the incline at 0.20 m/s^2.

1. Find the friction force impeding its notion?

2. What is the coefficient of kinetic friction?

*Are we dealing with Kinetic or Static Friction...this will effect coefficient correct?
*The equation coefficient*m*g*cos angle DOES NOT give you the friction force

  • Physics - ,

    Wb = m*g = 13kg * 9.8N/kg = 127.4 N. =
    Wt. of box.

    Fb = 127.4N @ 33 Deg. = Foce of box.
    Fp = 127.4*sin33 = 69.4 N. = Force parallel to incline.
    Fv = 127.4*cos33 = 106.8 N. = Force perpendicular to incline.

    1. Fn = Fp - Fk = ma
    69.4 - Fk = 13 * 0.20 = 2.6
    -Fk = 2.6 -69.4 = -72
    Fk = 72 N = Force of kinetic friction.

    2. u*Fv = Fk
    127.4u = 72
    u = 72 / 127.4 = 0.565 = Coefficient of
    kinetic friction.

  • Physics - ,

    Correction:

    -Fk = 2.6 - 69.4 = -66.8Fk = 66.8 N.

    2. u*Fv = Fk
    127.4u = 66.8
    u = 66.8 / 127.4 = 0.524 = uk.

  • Physics - ,

    OOPs!
    2. u*Fv = Fk
    1o6.8u = 66.8
    u = 66.8 / 106.8 = 0.625.

    NOTE: Fv = 106.8 N.

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