A body, constrained to move in the Y-direction is subjected to a force given by `vecF=(-2hati+15hatj+6hatk)N`. What is the work done by this force in moving the body a distance 10 m along the Y-axis

Options

A. (a)20 J
B. (b)150 J
C. (c)160 J
D. (d)190 J

Answer

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-by-Step Solution:

1. **Identify the Force Vector**:
The force vector is given as:
\[ \vec{F} = -2\hat{i} + 15\hat{j} + 6\hat{k} \text{ N} \]

2. **Identify the Displacement Vector**:
The body moves a distance of 10 m along the Y-axis. Therefore, the displacement vector can be represented as:
\[ \vec{s} = 10\hat{j} \text{ m} \]

3. **Calculate the Work Done**:
The work done \( W \) by the force when moving through a displacement is given by the dot product of the force vector and the displacement vector:
\[ W = \vec{F} \cdot \vec{s} \]

4. **Perform the Dot Product**:
To calculate the dot product, we use the formula:
\[ \vec{F} \cdot \vec{s} = (-2\hat{i} + 15\hat{j} + 6\hat{k}) \cdot (10\hat{j}) \]
This expands to:
\[ W = (-2\hat{i} \cdot 10\hat{j}) + (15\hat{j} \cdot 10\hat{j}) + (6\hat{k} \cdot 10\hat{j}) \]

5. **Evaluate Each Term**:
- The dot product of \(\hat{i}\) and \(\hat{j}\) is 0:
\[ -2\hat{i} \cdot 10\hat{j} = 0 \]
- The dot product of \(\hat{j}\) with itself is 1:
\[ 15\hat{j} \cdot 10\hat{j} = 150 \]
- The dot product of \(\hat{k}\) and \(\hat{j}\) is also 0:
\[ 6\hat{k} \cdot 10\hat{j} = 0 \]

6. **Combine the Results**:
Therefore, the total work done is:
\[ W = 0 + 150 + 0 = 150 \text{ J} \]

### Final Answer:
The work done by the force in moving the body a distance of 10 m along the Y-axis is:
\[ W = 150 \text{ J} \]