a 55.5 kg sled initially at rest, slides down a hill that is inclined at 36.0°. what is the magnitude of the…

Explanation:

Step1: Analyze forces perpendicular to the hill

The forces acting on the sled perpendicular to the hill are the normal force $n$ and the perpendicular - component of the gravitational force. The gravitational force is $F_g = mg$, where $m = 55.5$ kg and $g=9.8$ m/s². The perpendicular - component of the gravitational force to the hill is $F_{g\perp}=mg\cos\theta$.

Step2: Set up equilibrium equation

In the direction perpendicular to the hill, since there is no acceleration in that direction ($a_y = 0$), the net force $\sum F_y=0$. So, $n - mg\cos\theta=0$, and $n = mg\cos\theta$.

Step3: Substitute values

Substitute $m = 55.5$ kg, $g = 9.8$ m/s², and $\theta = 36.0^{\circ}$ into the formula. $\cos(36.0^{\circ})\approx0.809$. Then $n=55.5\times9.8\times0.809$. $n = 55.5\times9.8\times0.809=55.5\times7.9282\approx440$ N.

Answer:

440