a 5.00 kg box is on a ramp that is inclined at 21.0°. what is the y - component of the weight of the box…

Explanation:

Step1: Calculate the weight of the box

The weight of an object is given by $W = mg$, where $m = 5.00\ kg$ and $g= 9.8\ m/s^{2}$. So $W=5.00\times9.8 = 49\ N$.

Step2: Find the y - component of the weight

If the ramp is inclined at an angle $\theta = 21.0^{\circ}$, the y - component of the weight is $W_y=W\cos\theta$. Substituting $W = 49\ N$ and $\theta = 21.0^{\circ}$, we get $W_y = 49\times\cos(21.0^{\circ})$. Using a calculator, $\cos(21.0^{\circ})\approx0.934$, so $W_y=49\times0.934 = 45.766\approx45.8\ N$.

Answer:

$45.8$