Question

a (16.0\text{ kg}) sled initially at rest, slides down a frictionless hill that is inclined at (23.8^{circ}).

what is the acceleration of the sled as it slides down the hill?

(a = ? \text{ m/s}^2)

Explanation:

⚡ Using what you learned: newton's laws of motion

Step 1: Identify forces parallel to the incline

The only force acting along the frictionless incline is the component of gravity pulling the sled down the slope:

$$ F_{\parallel} = m \cdot g \cdot \sin(\theta) $$

Step 2: Apply Newton's Second Law

Set the net force equal to mass times acceleration:

$$ m \cdot a = m \cdot g \cdot \sin(\theta) $$

Divide both sides by mass \( m \):

$$ a = g \cdot \sin(\theta) $$

Step 3: Calculate the acceleration

Using \( g = 9.80 \text{ m/s}^2 \) and \( \theta = 23.8^\circ \):

$$ a = 9.80 \cdot \sin(23.8^\circ) $$

$$ a \approx 9.80 \cdot 0.4035 $$

$$ a \approx 3.95 \text{ m/s}^2 $$

Answer:

3.95