A force `F = (x hat i+2y hat j)`N is applied on an object of mass 10 kg, Force displaces the object from position A(1, 0) m to position B(3, 3) m then the work done by the force is (x and y are meter)

Options

A. 8 J
B. 5 J
C. 13 J
D. 16 J

Answer

To solve the problem of calculating the work done by the force \( \mathbf{F} = (x \hat{i} + 2y \hat{j}) \) N as it displaces an object from position \( A(1, 0) \) m to position \( B(3, 3) \) m, we will follow these steps:

### Step 1: Understand the Force and Displacement
The force is given as \( \mathbf{F} = x \hat{i} + 2y \hat{j} \). The displacement vector \( d\mathbf{s} \) can be expressed in terms of its components as \( d\mathbf{s} = dx \hat{i} + dy \hat{j} \).

### Step 2: Set Up the Work Done Integral
The work done \( W \) by a variable force is calculated using the integral:
\[ W = \int \mathbf{F} \cdot d\mathbf{s} \]
Substituting for \( \mathbf{F} \) and \( d\mathbf{s} \):
\[ W = \int (x \hat{i} + 2y \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \]

### Step 3: Calculate the Dot Product
Calculating the dot product:
\[ \mathbf{F} \cdot d\mathbf{s} = x \, dx + 2y \, dy \]
Thus, the work done becomes:
\[ W = \int (x \, dx + 2y \, dy) \]

### Step 4: Determine the Path of Integration
Since the force is variable, we need to express \( y \) in terms of \( x \) or vice versa. The path from \( A(1, 0) \) to \( B(3, 3) \) can be described as \( y = \frac{3}{2}(x - 1) \) (a straight line from A to B).

### Step 5: Express \( dy \) in Terms of \( dx \)
Differentiating \( y = \frac{3}{2}(x - 1) \):
\[ dy = \frac{3}{2} dx \]

### Step 6: Substitute \( y \) and \( dy \) into the Work Integral
Now we can substitute \( y \) and \( dy \) into the work integral:
\[ W = \int_{1}^{3} \left( x \, dx + 2 \left(\frac{3}{2}(x - 1)\right) \left(\frac{3}{2} dx\right) \right) \]
This simplifies to:
\[ W = \int_{1}^{3} \left( x \, dx + \frac{9}{2}(x - 1) \, dx \right) \]

### Step 7: Combine the Integrals
Combine the integrals:
\[ W = \int_{1}^{3} \left( x + \frac{9}{2}(x - 1) \right) dx = \int_{1}^{3} \left( x + \frac{9}{2}x - \frac{9}{2} \right) dx \]
\[ = \int_{1}^{3} \left( \frac{11}{2}x - \frac{9}{2} \right) dx \]

### Step 8: Evaluate the Integral
Now, evaluate the integral:
\[ W = \left[ \frac{11}{4}x^2 - \frac{9}{2}x \right]_{1}^{3} \]
Calculating the limits:
\[ W = \left( \frac{11}{4}(3^2) - \frac{9}{2}(3) \right) - \left( \frac{11}{4}(1^2) - \frac{9}{2}(1) \right) \]
\[ = \left( \frac{11 \cdot 9}{4} - \frac{27}{2} \right) - \left( \frac{11}{4} - \frac{9}{2} \right) \]
\[ = \left( \frac{99}{4} - \frac{54}{4} \right) - \left( \frac{11}{4} - \frac{18}{4} \right) \]
\[ = \frac{45}{4} - \left( \frac{-7}{4} \right) = \frac{45 + 7}{4} = \frac{52}{4} = 13 \, \text{J} \]

### Final Answer
The work done by the force is \( 13 \, \text{J} \).

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