A 35.2 kg wagon is towed up a hill inclined at 18.3◦ with respect to the horizontal. The tow rope is parallel to the incline and exerts a force of 125 N on the wagon. Assume that the wagon starts from rest at the bottom of the hill, and disregard friction.

Question

A 35.2 kg wagon is towed up a hill inclined at 18.3◦ with respect to the horizontal. The tow rope is parallel to the incline and exerts a force of 125 N on the wagon. Assume that the wagon starts from rest at the bottom of the hill, and disregard friction.
The acceleration of gravity is 9.81 m/s2 .
How fast is the wagon going after moving 75.4 m up the hill?
Answer in units of m/s

Henry · Accepted Answer

Fw = m*g = 35.2 * 9.8 = 345 N. = Force of the wagon.

Fp = 345*sin18.3 = 108.3 N. = Force parallel to the hill.

a = (Fex-Fp)/m = (125-108)/35.2 = 0.474
m/s^2.

V^2 = Vo^2 + 2a*d = 0 + 0.949*75.4 = 71.54
V = 8.46 m/s.