A 4-person bobsled team pushes the 330 kg sled down the 6.0 degree hill with a force of 1780 N. The coefficient of friction for ice and the runner is 0.135. How fast will the sled be moving at the end of the 40m push? Answer is in meters/second

Answer 1

M*g = 330 * 9.8 = 3234 N. = Wt. of sled.

Fp = 3234*sin6 = 38 N. = Force parallel
to the incline.

Fn = 3234*Cos6 = 3216 N. = Normal = Force perpendicular to the incline.

Fk = u*Fn = 0.135 * 3216 = 434.2 N. = Force of kinetic friction.

a = (Fap-Fk)/M = (1780-434)/330 = 4.08 m/s^2

V^2 = Vo^2 + 2a*d

Vo = 0
a = 4.08 m/s^2
d = 40 m.
Solve for V.