A toy cart of mass `sqrt(3)` kg is pulled by a force of 20N at an angle of `30^(@)` with the frictionless horizontal surface on which the cart is placed. The cart shall move on the surface with an acceleration .

Options

A. `0 m//s^(2)`
B. `1 m//s^(2)`
C. `10m//s^(2)`
D. `10sqrt( 3) m//s^(2)`

Answer

To solve the problem, we need to find the acceleration of a toy cart of mass \( \sqrt{3} \) kg that is being pulled by a force of 20 N at an angle of \( 30^\circ \) on a frictionless horizontal surface.

### Step-by-Step Solution:

1. **Identify the Forces:**
The force applied to the cart is \( F = 20 \, \text{N} \) at an angle of \( 30^\circ \) to the horizontal. We can resolve this force into its horizontal and vertical components.

2. **Resolve the Force:**
- The horizontal component of the force (\( F_x \)) can be calculated using:
\[ F_x = F \cdot \cos(\theta) = 20 \cdot \cos(30^\circ) \]
- The vertical component of the force (\( F_y \)) can be calculated using:
\[ F_y = F \cdot \sin(\theta) = 20 \cdot \sin(30^\circ) \]

3. **Calculate the Components:**
- Using the values of \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( \sin(30^\circ) = \frac{1}{2} \):
\[ F_x = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{N} \]
\[ F_y = 20 \cdot \frac{1}{2} = 10 \, \text{N} \]

4. **Apply Newton's Second Law:**
Since the surface is frictionless, we only need to consider the horizontal component of the force to find the acceleration. According to Newton's second law:
\[ F_x = m \cdot a \]
where \( m = \sqrt{3} \, \text{kg} \) and \( a \) is the acceleration we want to find.

5. **Substitute and Solve for Acceleration:**
\[ 10\sqrt{3} = \sqrt{3} \cdot a \]
To isolate \( a \), divide both sides by \( \sqrt{3} \):
\[ a = \frac{10\sqrt{3}}{\sqrt{3}} = 10 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the cart is \( 10 \, \text{m/s}^2 \).

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