Question

a (5.00\text{ kg}) box is on a ramp that is inclined at (21.0^{circ}).

what is the y-component of the weight of the box?

(w_y = ? \text{ n})

Explanation:

⚡ Using what you learned: Vectors in Plane and Space (components, magnitude, direction)

Step 1: Calculate total weight

The total weight \( w \) of the box is the force due to gravity acting straight down:

$$ w = m \cdot g $$

$$ w = 5.00 \text{ kg} \times 9.80 \text{ m/s}^2 = 49.0 \text{ N} $$

Step 2: Find the y-component perpendicular to the ramp

On an inclined plane tilted at an angle \( \theta \), the component of weight perpendicular to the ramp (the y-component) is given by:

$$ w_y = w \cdot \cos(\theta) $$

$$ w_y = 49.0 \text{ N} \cdot \cos(21.0^\circ) $$

$$ w_y \approx 49.0 \text{ N} \cdot 0.93358 \approx 45.7 \text{ N} $$

(Note: Depending on the coordinate system convention, the downward perpendicular component is sometimes written as negative, \( -45.7 \text{ N} \). If only the magnitude is required, it is \( 45.7 \text{ N} \).)

Answer:

-45.7 (or 45.7 if entering magnitude)