Asked by Ali
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.
Answers
Answered by
Henry
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.
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.
Answered by
William
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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