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.

Answer 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.

Answer 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.