A block of mass, 2 kg is kept on a smooth surface as shown. It is being applied by a force `F = 20 hat(i) + 10 hat(j)` Find the displacement of the block in 10 seconds.

Options

A. 500m
B. 10m
C. 100m
D. 50m

Answer

To solve the problem of finding the displacement of a block of mass 2 kg subjected to a force \( F = 20 \hat{i} + 10 \hat{j} \) over a time period of 10 seconds, we can follow these steps:

### Step 1: Identify the Force Components
The force \( F \) has two components:
- In the x-direction: \( F_x = 20 \, \text{N} \)
- In the y-direction: \( F_y = 10 \, \text{N} \)

### Step 2: Calculate the Acceleration
Using Newton's second law, we can find the acceleration \( a \) of the block. The formula is:
\[ F = m \cdot a \]
Where \( m \) is the mass of the block. Rearranging gives:
\[ a = \frac{F}{m} \]
Calculating the acceleration in each direction:
- For the x-direction:
\[ a_x = \frac{F_x}{m} = \frac{20 \, \text{N}}{2 \, \text{kg}} = 10 \, \text{m/s}^2 \]
- For the y-direction:
\[ a_y = \frac{F_y}{m} = \frac{10 \, \text{N}}{2 \, \text{kg}} = 5 \, \text{m/s}^2 \]

### Step 3: Use the Kinematic Equation for Displacement
The displacement \( s \) can be calculated using the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the block starts from rest, the initial velocity \( u = 0 \). Therefore, the equation simplifies to:
\[ s = \frac{1}{2} a t^2 \]

### Step 4: Calculate Displacement in Each Direction
Now, we will calculate the displacement in both the x and y directions over the time period of \( t = 10 \, \text{s} \).

- Displacement in the x-direction:
\[ s_x = \frac{1}{2} a_x t^2 = \frac{1}{2} \cdot 10 \, \text{m/s}^2 \cdot (10 \, \text{s})^2 = \frac{1}{2} \cdot 10 \cdot 100 = 500 \, \text{m} \]

- Displacement in the y-direction:
\[ s_y = \frac{1}{2} a_y t^2 = \frac{1}{2} \cdot 5 \, \text{m/s}^2 \cdot (10 \, \text{s})^2 = \frac{1}{2} \cdot 5 \cdot 100 = 250 \, \text{m} \]

### Step 5: Combine the Displacements
The total displacement vector \( \vec{s} \) is given by:
\[ \vec{s} = s_x \hat{i} + s_y \hat{j} = 500 \hat{i} + 250 \hat{j} \, \text{m} \]

### Final Answer
The displacement of the block in 10 seconds is:
\[ \vec{s} = 500 \hat{i} + 250 \hat{j} \, \text{m} \]

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