Asked by Vanessa
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?
Answers
Answered by
Henry
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 = ?.
Fk = u*Fn = 0.7 * 343 = 240 N. = Force of kinetic friction.
Fap-Fk = M*a.
275-240 = 35a, a = ?.
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