Two constant forces `vecF_1` and `vecF_2` act on a body of mass `8kg`. These forces displace the body from point `P(1, -2, 3)` to `Q` `(2, 3, 7)` in `2s` starting from rest. Force `vecF_1` is of magnitude `9N` and is acting along vector `(2hati-2hatj+hatk)`. Work done by the force `vecF_2` is

Options

A. (a) `80J`
B. (b) `-80J`
C. (c) `-180J`
D. (d) `180J`

Explanation

To solve the problem of finding the work done by the force \( \vec{F_2} \), we will follow these steps:

### Step 1: Identify Given Information
- Mass of the body, \( m = 8 \, \text{kg} \)
- Initial position \( P(1, -2, 3) \)
- Final position \( Q(2, 3, 7) \)
- Time taken, \( t = 2 \, \text{s} \)
- Magnitude of force \( \vec{F_1} = 9 \, \text{N} \) acting along the direction of \( (2\hat{i} - 2\hat{j} + \hat{k}) \)

### Step 2: Calculate the Displacement Vector
The displacement vector \( \vec{s} \) can be calculated as:
\[ \vec{s} = \vec{Q} - \vec{P} = (2\hat{i} + 3\hat{j} + 7\hat{k}) - (1\hat{i} - 2\hat{j} + 3\hat{k}) \]
Calculating this gives:
\[ \vec{s} = (2 - 1)\hat{i} + (3 + 2)\hat{j} + (7 - 3)\hat{k} = 1\hat{i} + 5\hat{j} + 4\hat{k} \]

### Step 3: Calculate the Acceleration
Since the body starts from rest, the initial velocity \( u = 0 \). We can use the equation of motion:
\[ \vec{s} = u t + \frac{1}{2} \vec{a} t^2 \]
Substituting \( u = 0 \):
\[ \vec{s} = \frac{1}{2} \vec{a} t^2 \]
Rearranging for \( \vec{a} \):
\[ \vec{a} = \frac{2\vec{s}}{t^2} = \frac{2(1\hat{i} + 5\hat{j} + 4\hat{k})}{(2)^2} = \frac{2(1\hat{i} + 5\hat{j} + 4\hat{k})}{4} = \frac{1}{2}(1\hat{i} + 5\hat{j} + 4\hat{k}) = \frac{1}{2}\hat{i} + \frac{5}{2}\hat{j} + 2\hat{k} \]

### Step 4: Calculate the Net Force
Using Newton's second law:
\[ \vec{F}_{\text{net}} = m \vec{a} = 8 \left( \frac{1}{2}\hat{i} + \frac{5}{2}\hat{j} + 2\hat{k} \right) = 4\hat{i} + 20\hat{j} + 16\hat{k} \]
Since \( \vec{F}_{\text{net}} = \vec{F_1} + \vec{F_2} \), we need to calculate \( \vec{F_1} \).

### Step 5: Calculate Force \( \vec{F_1} \)
The unit vector in the direction of \( \vec{F_1} \) is:
\[ \text{Magnitude of direction vector} = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = 3 \]
Thus, the unit vector is:
\[ \hat{F_1} = \frac{2\hat{i} - 2\hat{j} + \hat{k}}{3} \]
Now, \( \vec{F_1} \):
\[ \vec{F_1} = 9 \hat{F_1} = 9 \left( \frac{2\hat{i} - 2\hat{j} + \hat{k}}{3} \right) = 6\hat{i} - 6\hat{j} + 3\hat{k} \]

### Step 6: Calculate Force \( \vec{F_2} \)
Using \( \vec{F}_{\text{net}} = \vec{F_1} + \vec{F_2} \):
\[ \vec{F_2} = \vec{F}_{\text{net}} - \vec{F_1} = (4\hat{i} + 20\hat{j} + 16\hat{k}) - (6\hat{i} - 6\hat{j} + 3\hat{k}) = -2\hat{i} + 26\hat{j} + 13\hat{k} \]

### Step 7: Calculate Work Done by \( \vec{F_2} \)
Work done \( W \) is given by:
\[ W = \vec{F_2} \cdot \vec{s} = (-2\hat{i} + 26\hat{j} + 13\hat{k}) \cdot (1\hat{i} + 5\hat{j} + 4\hat{k}) \]
Calculating the dot product:
\[ W = (-2)(1) + (26)(5) + (13)(4) = -2 + 130 + 52 = 180 \, \text{J} \]

### Final Answer
The work done by the force \( \vec{F_2} \) is \( 180 \, \text{J} \).

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