Question

a block weighing 10. newtons rests on an inclined plane, as shown in the diagram below. the magnitude of the component of the blocks weight perpendicular to the plane is closest to (1) 5.0 n (2) 5.8 n (3) 8.7 n (4) 10. n

Explanation:

Step1: Identify the formula

The component of the weight perpendicular to the inclined - plane is given by $F_{perp}=F_g\cos\theta$, where $F_g$ is the weight of the block and $\theta$ is the angle of the inclined plane.

Step2: Substitute the values

We know that $F_g = 10\ N$ and $\theta=30^{\circ}$, and $\cos30^{\circ}=\frac{\sqrt{3}}{2}\approx0.866$. Then $F_{perp}=10\times\cos30^{\circ}=10\times0.866 = 8.66\ N\approx8.7\ N$.

Answer:

(3) $8.7\ N$