Question

A crate of mass 50kg is pulled up a rough inclined plane ,inclined at an angle of 60 degree to the horizontal.determine the total work done in moving the crate to the top of the inclined 200 meter hight.(show workin).
Wc = m*g = 50kg * 9.8N/kg = 490 N = Wt.
of crate.

Fc = 490N @ 60o = Force of crate.
Fp = 490*sin60 = 424.4 N. = Force
parallel to incline.
Fv = 490*cos60 = 245 N. = Force perpendicular to the incline.

Ff = u*Fv = u*245 = Force of friction.

Fap-Fp-Ff = m*a
Fap-424.4-u*245 = m*0 = 0
Fap = 424.4 + 245u. = Force applied.

L = 200m/sin60 = 231 m. = Length of
incline.

Work = Fap * L=(424.4+245u) * 231.

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