Under a force `(10hat(i)-3hat(j)+6hat(k))` Newton a body of mass 5 kg moves from position `(6hat(i)+5hat(j)-3hat(k))` m to position `(10hat(i)-2hat(j)+7hat(k))` m. Deduce the work done.

Options

A. `-121 J`
B. 121 J
C. 242 J
D. 0 J

Explanation

To find the work done by the force on the body as it moves from one position to another, we can follow these steps:

### Step 1: Determine the Displacement Vector
The displacement vector \(\vec{s}\) can be calculated by subtracting the initial position vector \(\vec{s_i}\) from the final position vector \(\vec{s_f}\).

Given:
- Initial position \(\vec{s_i} = 6\hat{i} + 5\hat{j} - 3\hat{k}\)
- Final position \(\vec{s_f} = 10\hat{i} - 2\hat{j} + 7\hat{k}\)

\[ \vec{s} = \vec{s_f} - \vec{s_i} = (10\hat{i} - 2\hat{j} + 7\hat{k}) - (6\hat{i} + 5\hat{j} - 3\hat{k}) \]

Calculating the components:
\[ \vec{s} = (10 - 6)\hat{i} + (-2 - 5)\hat{j} + (7 + 3)\hat{k} = 4\hat{i} - 7\hat{j} + 10\hat{k} \]

### Step 2: Calculate the Work Done
The work done \(W\) by the force \(\vec{F}\) on the displacement \(\vec{s}\) can be calculated using the dot product:

Given:
- Force \(\vec{F} = 10\hat{i} - 3\hat{j} + 6\hat{k}\)

The work done is given by:
\[ W = \vec{F} \cdot \vec{s} \]

Calculating the dot product:
\[ W = (10\hat{i} - 3\hat{j} + 6\hat{k}) \cdot (4\hat{i} - 7\hat{j} + 10\hat{k}) \]

Calculating each component:
\[ W = (10 \cdot 4) + (-3 \cdot -7) + (6 \cdot 10) = 40 + 21 + 60 \]

Adding these values together:
\[ W = 121 \text{ Joules} \]

### Final Answer
The work done by the force is \(121\) Joules.

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