Question

a 1520 kg car in motion rolls down a frictionless hill that is inclined at (14.4^{circ}).

what is the acceleration of the car?

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

Explanation:

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

Step 1: Identify the forces acting parallel to the incline

For an object on a frictionless incline of angle \(\theta\), the component of gravity acting parallel to the slope is:

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

Step 2: Apply Newton's Second Law

Set the net force parallel to the incline equal to \(m a\):

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

Divide both sides by mass \(m\):

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

Step 3: Calculate the acceleration

Using \(g = 9.8\text{ m/s}^2\) and \(\theta = 14.4^{\circ}\):

$$ a = 9.8 \cdot \sin(14.4^{\circ}) $$

$$ a \approx 9.8 \cdot 0.2487 $$

$$ a \approx 2.44\text{ m/s}^2 $$

(Note: If using \(g = 9.81\text{ m/s}^2\), \(a \approx 9.81 \cdot 0.2487 \approx 2.44\text{ m/s}^2\))

Answer:

2.44