A boy pushes a box of mass 2 kg with a force `vecF=(20hati+10hatj)` Non a frictionless surface. If the box was initially at rest, then `"________"` m is displacement along the x-axis after 10 s.

Answer

To solve the problem step by step, we will follow these procedures:

### Step 1: Identify the Given Information
- Mass of the box (m) = 2 kg
- Applied force (F) = \(20 \hat{i} + 10 \hat{j}\) N
- Initial velocity (u) = 0 m/s (the box is initially at rest)
- Time (t) = 10 s

### Step 2: Calculate the Acceleration
Using Newton's second law, the acceleration (a) can be calculated using the formula:
\[ a = \frac{F}{m} \]
Substituting the values:
\[ F = 20 \hat{i} + 10 \hat{j} \quad \text{(we only need the x-component for displacement along the x-axis)} \]
The x-component of the force \(F_x = 20 \, \text{N}\).

Now, calculate the acceleration in the x-direction:
\[ a_x = \frac{F_x}{m} = \frac{20 \, \text{N}}{2 \, \text{kg}} = 10 \, \text{m/s}^2 \]

### Step 3: Use the Kinematic Equation for Displacement
The formula for displacement (S) under constant acceleration is:
\[ S = ut + \frac{1}{2} a t^2 \]
Since the box starts from rest, the initial velocity \(u = 0\). Therefore, the equation simplifies to:
\[ S = 0 \cdot t + \frac{1}{2} a t^2 = \frac{1}{2} a t^2 \]

### Step 4: Substitute the Values
Now substitute the values of acceleration and time into the equation:
\[ S = \frac{1}{2} \cdot 10 \, \text{m/s}^2 \cdot (10 \, \text{s})^2 \]
\[ S = \frac{1}{2} \cdot 10 \cdot 100 = 500 \, \text{m} \]

### Conclusion
The displacement along the x-axis after 10 seconds is **500 m**.

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