A body constrained to move along y-axis is subjected to a constant force `vecF= -hati+2hatj+3hatk N`. The work done by this force in moving the body a distance of 4 m along y-axis is

Options

A. 4J
B. 8 J
C. 12J
D. 24 J

Explanation

To solve the problem of calculating the work done by the force on the body moving along the y-axis, we can follow these steps:

### Step 1: Identify the Force and Displacement
The force acting on the body is given as:
\[ \vec{F} = -\hat{i} + 2\hat{j} + 3\hat{k} \text{ N} \]
The displacement of the body is along the y-axis for a distance of 4 m, which can be expressed as:
\[ \vec{d} = 4\hat{j} \text{ m} \]

### Step 2: Use the Work Done Formula
The work done \( W \) by a force when moving an object is given by the dot product of the force vector and the displacement vector:
\[ W = \vec{F} \cdot \vec{d} \]

### Step 3: Calculate the Dot Product
Now, we will calculate the dot product:
\[ \vec{F} \cdot \vec{d} = (-\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (4\hat{j}) \]
Using the properties of the dot product, we can expand this:
\[ = (-\hat{i} \cdot 4\hat{j}) + (2\hat{j} \cdot 4\hat{j}) + (3\hat{k} \cdot 4\hat{j}) \]
Since the dot product of perpendicular vectors is zero, we have:
\[ -\hat{i} \cdot \hat{j} = 0 \quad \text{and} \quad 3\hat{k} \cdot \hat{j} = 0 \]
Thus, we only need to calculate the second term:
\[ = 0 + (2 \cdot 4) + 0 = 8 \text{ J} \]

### Step 4: Conclusion
The work done by the force in moving the body a distance of 4 m along the y-axis is:
\[ W = 8 \text{ Joules} \]

### Final Answer
The answer is \( 8 \text{ J} \).

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