Question

a (5.20\text{ kg}) ball is on a hill that is inclined at (15.0^{circ}).

what is the x-component of the weight of the ball?

(w_x = ? \text{ n})

Explanation:

🆕 New Concept Discovered: Forces on an Inclined Plane
Splitting gravity into parallel and perpendicular components

Step 1: Calculate the total weight of the ball

The weight \( w \) of an object is the force of gravity acting on its mass \( m \). We use the acceleration due to gravity \( g \approx 9.80\text{ m/s}^2 \):

$$ w = m \cdot g $$

$$ w = 5.20\text{ kg} \times 9.80\text{ m/s}^2 = 50.96\text{ N} $$

Step 2: Find the x-component of the weight

When an object is on an inclined plane, we set up a coordinate system where the x-axis is parallel to the slope (pointing down the hill) and the y-axis is perpendicular to the slope.

The component of gravity pulling the object down the slope (the x-component, \( w_x \)) is related to the incline angle \(\theta\) by the sine function:

$$ w_x = w \cdot \sin(\theta) $$

Substitute the values into the formula:

$$ w_x = 50.96\text{ N} \times \sin(15.0^\circ) $$

$$ w_x \approx 50.96 \times 0.2588 = 13.19\text{ N} $$

Rounding to three significant figures (matching the given values \(5.20\text{ kg}\) and \(15.0^\circ\)):

$$ w_x \approx 13.2\text{ N} $$

Answer:

13.2