Asked by Rosie
A wagon is being pushed by a small child along a sidewalk. If the wagon has a mass of 8kg, a coefficient of friction of 0.3 and is accelerating at 0.2m/s what force is the child applying
Answers
Answered by
Henry
Wt. = m*g = 8kg * 9.8N/kg = 78.4 N.
Fw = 78.4N.[0o] = Force of wagon.
Fp = 78.4*sin(0) = 0 = Force parallel to sidewalk.
Fv = 78.4*cos(0) = 78.4 N. = Force
perpendicular to sidewalk.
Fk = u*Fv = 0.3*78.4 = 23.52 N = Force of kinetic friction.
Fap-Fp-Fk = m*a.
Fap-0-23.52 = 8*0.2
Fap-23.52 = 1.6
Fap = 1.6 + 23.52 = 25.1 N. = Force applied.
Fw = 78.4N.[0o] = Force of wagon.
Fp = 78.4*sin(0) = 0 = Force parallel to sidewalk.
Fv = 78.4*cos(0) = 78.4 N. = Force
perpendicular to sidewalk.
Fk = u*Fv = 0.3*78.4 = 23.52 N = Force of kinetic friction.
Fap-Fp-Fk = m*a.
Fap-0-23.52 = 8*0.2
Fap-23.52 = 1.6
Fap = 1.6 + 23.52 = 25.1 N. = Force applied.
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