QuestionAs part of the Avian Olympics, an owl ( $m=4.49 \mathrm{~kg}$ ) slides down a semi-circular ramp of radius $R=3.7 \mathrm{~m}$. The owl starts at the top of the ramp at rest. When the owl is an angle of $\theta=32.9^{\circ}$, what is the normal force exerted by the ramp on the owl? a. 47.9 N cross out b. 57.1 N cross out c. 44.0 N cross out d. 71.9 N cross out e. 73.9 N cross out

Studdy Solution
Substitute the known values: $$N = 4.49 \times 9.81 \times (3\cos(32.9^\circ) - 2)$$
Calculate $\cos(32.9^\circ)$ and then $N$.
$$\cos(32.9^\circ) \approx 0.8387$$
$$N \approx 4.49 \times 9.81 \times (3 \times 0.8387 - 2)$$
$$N \approx 4.49 \times 9.81 \times 0.5161$$
$$N \approx 22.7 \, \text{N}$$
The normal force exerted by the ramp on the owl is approximately:
$$\boxed{22.7 \, \text{N}}$$