Question

question 19 of 25 frieda stands on a hill with a slope of 62°. if her mass is 49 kg, what is the magnitude of the normal force acting on her? a. 424 n b. 512 n c. 226 n d. 480 n

Explanation:

Step1: Resolve weight into components

The weight of Frieda is $W = mg$, where $m = 49$ kg and $g=9.8$ m/s². The normal force $N$ on an inclined - plane is related to the component of the weight perpendicular to the plane. The weight $W = 49\times9.8=480.2$ N. The component of the weight perpendicular to the inclined plane is $N = mg\cos\theta$, where $\theta = 62^{\circ}$.

Step2: Calculate the normal force

$N=49\times9.8\times\cos62^{\circ}$. We know that $\cos62^{\circ}\approx0.4695$. So $N = 49\times9.8\times0.4695=480.2\times0.4695\approx226$ N.

Answer:

C. 226 N