Asked by tim
If a pull of 350 N accelerates a 38-kg child on ice skates at a rate of 7.3 \frac{m}{s^2}, what is the frictional force acting on the skate?
Answers
Answered by
Henry
Fap-Fk = m*a
350-Fk = 38*7.3 = 277.4
-Fk = 277.4-350 = -72.6
Fk = 72.6 N. = Force of kinetic friction.
350-Fk = 38*7.3 = 277.4
-Fk = 277.4-350 = -72.6
Fk = 72.6 N. = Force of kinetic friction.
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