Question

A sled and rider have a total mass of 49.6 kg. They are on a snowy hill accelerating at 0.8g. The coefficient of kinetic friction between the sled and the snow is 0.22. What is the angle of the hill's slope measured upward from the horizontal? You may find a spreadsheet program helpful in answering this question.

Answer 1

force acting down the slope = mg sin

friction force acting up the plane = umg cos

acceleration= Net force / mass

= (mg sin - umg cos ) / m

= g sin - ug cos

0.8g = g(sin - 0.22*cos )

0.8 = sin - 0.22 cos

sin = 0.8 + 0.22 cos

squaring both sides we have

sin^2 = (0.8 + 0.22 cos )^2

(1- cos^2 ) = 0.64 + 0.0484 cos^2 + 0.352 cos

1.0484 cos^2 + 0.352 cos - 0.36 = 0

cos   = - 0.352 + sqrt(0.352^2 + 4*0.36*1.0484) / 2*1.0484

= 63.8^o

so the angle of the hill's slope is 63.8^o