Question
An object with a mass of 3.7 kg slides in the positive x direction across a horizontal surface with a coefficient of kinetic friction of 0.24. During its slide a force of 364 Newtons acts on it in the positive x direction while a force of 44 Newtons acts on it in the negative x direction. If the object gains 548 Joules of kinetic energy during its slide, how far did the object slide in meters?
Help me solve please .
Help me solve please .
Answers
Wo = mg = 3.7kg * 9.8N/kg = 36.26N. =
Weight of object.
Fo = 36.26N@0deg.
Fp=36.26sin(0) =0=Force perpendicular to surface.
Fv = 36.26cos(0) = 36.26N. = Force
perpendicular to surface.
Ff = u*Fv = 0.24 * 36.26 = 8.70N. =
Force of friction.
Fn = (F1 + F2) - Fp - Ff,
Fn = (324 - 44) - 8.70 = 271.3N.
Fn = ma,
a = Fn / m = 271.3 / 3.7 = 73.3m/s^2.
KE = 0.5mV^2 = 548J.
1.85V^2 = 548,
V^2 = 296.2,
V = 17.2m/s.
d = (Vf^2 - Vo^2) / 2a,
d = ((17.2)^2 - 0) / 146.6 = 2.02m.
Weight of object.
Fo = 36.26N@0deg.
Fp=36.26sin(0) =0=Force perpendicular to surface.
Fv = 36.26cos(0) = 36.26N. = Force
perpendicular to surface.
Ff = u*Fv = 0.24 * 36.26 = 8.70N. =
Force of friction.
Fn = (F1 + F2) - Fp - Ff,
Fn = (324 - 44) - 8.70 = 271.3N.
Fn = ma,
a = Fn / m = 271.3 / 3.7 = 73.3m/s^2.
KE = 0.5mV^2 = 548J.
1.85V^2 = 548,
V^2 = 296.2,
V = 17.2m/s.
d = (Vf^2 - Vo^2) / 2a,
d = ((17.2)^2 - 0) / 146.6 = 2.02m.