What acceleration would a 35 KG box have if we pushed it with a force of 275 N across a surface that has a coefficient of friction of .7?

Answer 1

M*g = 35 * 9.8 = 343 N. = Wt. of box = Normal force(Fn ).

Fk = u*Fn = 0.7 * 343 = 240 N. = Force of kinetic friction.

Fap-Fk = M*a.
275-240 = 35a, a = ?.