Asked by Mandeep
A man pulls a 25kg sled a distance of 80 meters across a flat, snowy surface. He holds the handle of the sled at an angle of 30degrees with the ground. How much work does he do on the sled against the force of friction?(Assume the coefficient of sliding friction between the sled and the snow to be .1, and cos30=.87).
Answers
Answered by
Henry
Ws = m*g = 25kg *9.8N/kg = 245 N. = Wt of sled.
Fs = 245N @ 0o. = Force of sled
Fp = 245*sin(0) = 0 = Force parallel to
surface.
Fv = 245*cos(0) = 245 N. = Force perpendicular to surface.
Fn = Fap*cos30-Fp-0.1(Fv-Fap*sin30) = ma
0.87Fap-0-0.1(245-0.5Fap) = 25*0 = 0
0.87Fap - 24.5 + 0.05Fap = 0
0.92Fap - 24.5 = 0
0.92Fap = 24.5
Fap = 24.5 / 0.92 = 26.6 N. = Force applied.
Work = Fap*cos30 * d
Work = 26.6*cos30 * 80 = 1851.4 Joules.
Fs = 245N @ 0o. = Force of sled
Fp = 245*sin(0) = 0 = Force parallel to
surface.
Fv = 245*cos(0) = 245 N. = Force perpendicular to surface.
Fn = Fap*cos30-Fp-0.1(Fv-Fap*sin30) = ma
0.87Fap-0-0.1(245-0.5Fap) = 25*0 = 0
0.87Fap - 24.5 + 0.05Fap = 0
0.92Fap - 24.5 = 0
0.92Fap = 24.5
Fap = 24.5 / 0.92 = 26.6 N. = Force applied.
Work = Fap*cos30 * d
Work = 26.6*cos30 * 80 = 1851.4 Joules.
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