Two constant forces `vec(F)_(1)` and `vec(F)_(2)` acts on a body of mass 8 kg. These forces displaces the body from point `P(1, -2,3)` to `Q(2,3,7)` in 2s starting from rest. Force `vec(F)_(1)` is of magnitude 9 N and is acting along vector `(2hat(i)-2hat(j)+hat(k))` . Work done by the force `vec(F)_(2)` is :-

Options

A. `80 J`
B. `-80 J`
C. `-180 J`
D. `180 J`

Explanation

To solve the problem step by step, we need to find the work done by the force \( \vec{F}_2 \) acting on a body of mass 8 kg, which is displaced from point \( P(1, -2, 3) \) to point \( Q(2, 3, 7) \) in 2 seconds.

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

### Step 2: Calculate the Magnitude of the Displacement
The magnitude of the displacement \( |\vec{PQ}| \) is given by:
\[ |\vec{PQ}| = \sqrt{(1)^2 + (5)^2 + (4)^2} = \sqrt{1 + 25 + 16} = \sqrt{42} \]

### Step 3: Determine the Force \( \vec{F}_1 \)
The force \( \vec{F}_1 \) has a magnitude of 9 N and acts along the direction of the vector \( (2\hat{i} - 2\hat{j} + \hat{k}) \). First, we need to normalize this vector:
\[ \text{Magnitude of direction vector} = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]
Thus, the unit vector in the direction of \( \vec{F}_1 \) is:
\[ \hat{n} = \frac{2\hat{i} - 2\hat{j} + \hat{k}}{3} \]
Now, we can find \( \vec{F}_1 \):
\[ \vec{F}_1 = 9 \cdot \hat{n} = 9 \cdot \left(\frac{2\hat{i} - 2\hat{j} + \hat{k}}{3}\right) = 6\hat{i} - 6\hat{j} + 3\hat{k} \]

### Step 4: Calculate the Net Force
Let \( \vec{F}_2 = F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k} \). The net force \( \vec{F}_{net} \) acting on the body is:
\[ \vec{F}_{net} = \vec{F}_1 + \vec{F}_2 \]

### Step 5: Use Newton's Second Law to Find Acceleration
Using Newton's second law, we have:
\[ \vec{F}_{net} = m \vec{a} \]
Where \( m = 8 \, \text{kg} \). The acceleration \( \vec{a} \) can be calculated using the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Since the body starts from rest, \( u = 0 \):
\[ |\vec{PQ}| = \frac{1}{2} a (2^2) \implies \sqrt{42} = 2a \implies a = \frac{\sqrt{42}}{2} \]

### Step 6: Calculate the Work Done by the Net Force
The work done by the net force is given by:
\[ W_{net} = \vec{F}_{net} \cdot \vec{PQ} \]
Using \( \vec{F}_{net} = m \vec{a} = 8 \cdot \frac{\sqrt{42}}{2} = 4\sqrt{42} \).

### Step 7: Calculate Work Done by \( \vec{F}_1 \)
The work done by \( \vec{F}_1 \) is:
\[ W_1 = \vec{F}_1 \cdot \vec{PQ} = (6\hat{i} - 6\hat{j} + 3\hat{k}) \cdot (\hat{i} + 5\hat{j} + 4\hat{k}) = 6(1) + (-6)(5) + 3(4) = 6 - 30 + 12 = -12 \, \text{J} \]

### Step 8: Calculate Work Done by \( \vec{F}_2 \)
Using the relationship:
\[ W_{net} = W_1 + W_2 \implies W_2 = W_{net} - W_1 \]
The net work done can be calculated as:
\[ W_{net} = F_{net} \cdot \text{displacement} = (8 \cdot \frac{\sqrt{42}}{2}) \cdot \sqrt{42} = 4 \cdot 42 = 168 \, \text{J} \]
Thus:
\[ W_2 = 168 - (-12) = 168 + 12 = 180 \, \text{J} \]

### Final Answer
The work done by the force \( \vec{F}_2 \) is:
\[ \boxed{180 \, \text{J}} \]