Question

a 2500 kg car initially sitting at rest rolls down a frictionless hill that is inclined at \\(37.2^{\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 forces parallel to the incline

The component of gravity pulling the car down the frictionless incline is:

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

Step 2: Apply Newton's Second Law

Since there is no friction, the net force is equal to this gravitational component:

$$ F_{\text{net}} = m \cdot a $$

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

The mass \( m \) cancels out from both sides:

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

Step 3: Calculate the acceleration

Using \( g = 9.8 \text{ m/s}^2 \) and \( \theta = 37.2^\circ \):

$$ a = 9.8 \cdot \sin(37.2^\circ) $$

$$ a \approx 9.8 \cdot 0.6046 $$

$$ a \approx 5.93 \text{ m/s}^2 $$

(Note: If using \( g = 9.81 \text{ m/s}^2 \), \( a = 9.81 \cdot \sin(37.2^\circ) \approx 5.93 \text{ m/s}^2 \); if using \( g = 9.8 \text{ m/s}^2 \), \( a \approx 5.9 \text{ m/s}^2 \))

Answer:

5.93