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A 60 kg is pushed 12m across a horizontal floor by a horizontal force of 200N. The coefficient of kinetic friction is 0.3. How much work went into overcoming friction and how much work went into accelerating the box.

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2 answers
  1. Wc = M*g = 60 * 9.8 = 588 N. = Normal force, Fn.

    Fk = u*Fn = 0.3 * 588 = 176.4 N.
    Work = Fk*d = 176.4 * 12 = 2117 J.

    Fap-Fk = M*a.
    200-176.4 = 60*a.
    60a = 23.6.
    a = 0.393 m/s^2.

    Work = M*a * d = 60*0.393 * 12 = 283 J.

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  2. A 60kg box is pushed 12M acrosse horizontal floor by a horizontal force of 200N. The coefficient of kinetie friction is 0.3. How much work went in to over coming friction. How much in to acceleration the box.

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