Posted by **Amanda ** on Thursday, October 27, 2011 at 2:45pm.

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 .

- PHYSICS PLEASE HELP !!!!! -
**Henry**, Friday, October 28, 2011 at 9:59pm
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.

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