A force`F=(3thati + 5hatj)N` acts on a body due to which its displacement varies as `S=(2t^(2)hat-5hatj)m`. Work done by these force in `2 s` is .

Options

A. `32 J`
B. `24 J`
C. `46 J`
D. `20 J`

Explanation

To find the work done by the force \( F = (3t \hat{i} + 5 \hat{j}) \, \text{N} \) on a body with displacement \( S = (2t^2 \hat{i} - 5 \hat{j}) \, \text{m} \) over a time interval of \( 2 \, \text{s} \), we will follow these steps:

### Step 1: Find the expression for \( ds \)
The displacement \( S \) is given as:
\[ S = 2t^2 \hat{i} - 5 \hat{j} \]
To find \( ds \), we differentiate \( S \) with respect to time \( t \):
\[ \frac{dS}{dt} = \frac{d}{dt}(2t^2 \hat{i} - 5 \hat{j}) = (4t \hat{i} + 0 \hat{j}) = 4t \hat{i} \]
Thus,
\[ ds = 4t \hat{i} \, dt \]

### Step 2: Write the expression for work done
The work done \( W \) by a variable force is given by the integral:
\[ W = \int_{0}^{T} F \cdot ds \]
Substituting \( F \) and \( ds \):
\[ W = \int_{0}^{2} (3t \hat{i} + 5 \hat{j}) \cdot (4t \hat{i}) \, dt \]

### Step 3: Calculate the dot product
Now we compute the dot product:
\[ F \cdot ds = (3t \hat{i} + 5 \hat{j}) \cdot (4t \hat{i}) = 3t \cdot 4t + 5 \cdot 0 = 12t^2 \]

### Step 4: Set up the integral
Now we can set up the integral for work done:
\[ W = \int_{0}^{2} 12t^2 \, dt \]

### Step 5: Evaluate the integral
Calculating the integral:
\[ W = 12 \int_{0}^{2} t^2 \, dt = 12 \left[ \frac{t^3}{3} \right]_{0}^{2} = 12 \left[ \frac{2^3}{3} - \frac{0^3}{3} \right] = 12 \left[ \frac{8}{3} \right] = \frac{96}{3} = 32 \, \text{J} \]

### Final Answer
The work done by the force in \( 2 \, \text{s} \) is:
\[ \boxed{32 \, \text{J}} \]