Question: A 2.0kg toolbox is released from rest on a frictionless ramp that extends from the back of a truck, the ramp has an angle of 30 degrees, calculate the acceleration of the toolbox down the ramp
1. **State the problem:**
Calculate the acceleration of a 2.0kg toolbox sliding down a frictionless ramp inclined at an angle of $30^\circ$.
2. **Formula and rules:**
The acceleration $a$ of an object sliding down a frictionless incline is given by:
$$a = g \sin(\theta)$$
where $g = 9.8$ m/s$^2$ is the acceleration due to gravity and $\theta$ is the angle of the incline.
3. **Apply the formula:**
Given $\theta = 30^\circ$, substitute into the formula:
$$a = 9.8 \times \sin(30^\circ)$$
4. **Calculate $\sin(30^\circ)$:**
$$\sin(30^\circ) = 0.5$$
5. **Calculate acceleration:**
$$a = 9.8 \times 0.5 = 4.9$$
6. **Interpretation:**
The toolbox accelerates down the ramp at $4.9$ m/s$^2$ due to gravity component along the incline.
**Final answer:**
$$\boxed{4.9 \text{ m/s}^2}$$