A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providing a steady force of 131 N that is parallel to the motion of the sled. The combined mass of the girl and the sled is 59.9 kg, and the coefficient of kinetic friction between the snow and the runners of the sled is 0.217. How much time is required for the sled to travel down a 243-m slope, starting from rest?

Question

I found the acceleration to be 4.28m/s^2. I plugged it into the x=vot+1/2at^2 to find the time, but it turns out to be incorrect. Is my acceleration not correct? Please help, thanks

Answer 1

M*g = 59.9 * 9.8 = 587 N. = Wt. of girl
and sled.

Fp = 587*sin30 + 131 = 424.5 N. = Force
parallel to the incline.

Fn = 587*Cos30 = 508.4 N. = Normal = Force perpendicular to the incline.

Fk = u*Fn = 0.217 * 508.4 = 110.3 N. =
Force of kinetic friction.

a=(Fp-Fk)/M = (424.5-110.3)/59.9 = 5.25
m/s^2.

d = 0.5*a*t^2 = 243 m.
0.5*5.25t^2 = 243
t^2 = 92.57
t = 9.62 s.

Answer 2

what about the force in the X direction coming from the weight of the girl and the sled