Question

a 60.0 kg box initially at rest, slides down a frictionless ramp that is inclined at 24.5^{\circ}.

what is the acceleration of the box as it slides down the ramp?

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

Explanation:

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

Step 1: Identify forces parallel to the ramp

The only force acting along the direction of the incline is the component of gravity pulling the box down the ramp:

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

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 \( g = 9.8 \text{ m/s}^2 \) and \( \theta = 24.5^\circ \):

$$ a = 9.8 \cdot \sin(24.5^\circ) $$

$$ a \approx 9.8 \cdot 0.4147 $$

$$ a \approx 4.06 \text{ m/s}^2 $$

Answer:

4.06