A bodys constrained to more in the Y-direction ,Is subject to a force `vecF=(-2hati+15hatj+6hatk)` N What is the work done by force in moving the body through a distance of 10 m along the Y-axis ?

Options

A. 20 J
B. 150 J
C. 160 J
D. 190 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 Vector
The force vector is given as:
\[ \vec{F} = -2 \hat{i} + 15 \hat{j} + 6 \hat{k} \text{ N} \]

### Step 2: Identify the Displacement Vector
Since the body is constrained to move only in the Y-direction, the displacement vector for a distance of 10 m along the Y-axis is:
\[ \vec{d} = 10 \hat{j} \text{ m} \]

### Step 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:
\[ W = \vec{F} \cdot \vec{d} \]

### Step 4: Perform the Dot Product
To compute the dot product:
\[ \vec{F} \cdot \vec{d} = (-2 \hat{i} + 15 \hat{j} + 6 \hat{k}) \cdot (10 \hat{j}) \]
This can be expanded as:
\[ W = (-2 \hat{i} \cdot 10 \hat{j}) + (15 \hat{j} \cdot 10 \hat{j}) + (6 \hat{k} \cdot 10 \hat{j}) \]
Since the dot product of perpendicular vectors is zero:
\[ W = 0 + (15 \times 10) + 0 = 150 \text{ J} \]

### Step 5: Conclusion
Thus, the work done by the force in moving the body through a distance of 10 m along the Y-axis is:
\[ W = 150 \text{ J} \]