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 the toy cart being pulled by a force at an angle. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Mass of the toy cart, \( m = \sqrt{3} \) kg
- Pulling force, \( F = 20 \) N
- Angle of the force with the horizontal, \( \theta = 30^\circ \)

### Step 2: Resolve the force into horizontal and vertical components
The force can be resolved into two components:
- Horizontal component \( F_x = F \cdot \cos(\theta) \)
- Vertical component \( F_y = F \cdot \sin(\theta) \)

Calculating the horizontal component:
\[ F_x = 20 \cdot \cos(30^\circ) \]
Using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ F_x = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \text{ N} \]

### Step 3: Apply Newton's second law
According to Newton's second law, the acceleration \( a \) can be calculated using the formula:
\[ F_x = m \cdot a \]
Substituting the values:
\[ 10\sqrt{3} = \sqrt{3} \cdot a \]

### Step 4: Solve for acceleration \( a \)
To find \( a \), we divide both sides by \( \sqrt{3} \):
\[ a = \frac{10\sqrt{3}}{\sqrt{3}} = 10 \text{ m/s}^2 \]

### Conclusion
The acceleration of the toy cart is \( 10 \text{ m/s}^2 \).

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