The work done by a force `vec(F)=-5hat(k)` as its point of application moves from the point `(1,1,1)` to the origin is equal to `…………………` .

Explanation

To find the work done by the force \(\vec{F} = -5\hat{k}\) as its point of application moves from the point \((1, 1, 1)\) to the origin \((0, 0, 0)\), we can follow these steps:

### Step 1: Understand the Work Done Formula
The work done \(W\) by a force along a path is given by the integral:
\[ W = \int \vec{F} \cdot d\vec{s} \]
where \(\vec{F}\) is the force vector and \(d\vec{s}\) is the differential displacement vector.

### Step 2: Define the Path of Motion
The point of application moves from \((1, 1, 1)\) to \((0, 0, 0)\). We can express the displacement vector \(d\vec{s}\) in terms of its components:
\[ d\vec{s} = dx\hat{i} + dy\hat{j} + dz\hat{k} \]

### Step 3: Identify the Force Vector
The force vector is given as:
\[ \vec{F} = -5\hat{k} \]
This means that the force is acting in the negative z-direction.

### Step 4: Set Up the Integral
Since the force only has a component in the z-direction, we can simplify the work done calculation by focusing on the z-component of the displacement. The change in z as we move from \((1, 1, 1)\) to \((0, 0, 0)\) is from \(z = 1\) to \(z = 0\).

Thus, we can express the work done as:
\[ W = \int_{1}^{0} \vec{F} \cdot d\vec{s} = \int_{1}^{0} (-5\hat{k}) \cdot (0\hat{i} + 0\hat{j} + dz\hat{k}) \]
This simplifies to:
\[ W = \int_{1}^{0} -5 \, dz \]

### Step 5: Evaluate the Integral
Now we can evaluate the integral:
\[ W = -5 \int_{1}^{0} dz = -5 [z]_{1}^{0} = -5 (0 - 1) = -5 \times (-1) = 5 \]

### Conclusion
The work done by the force as it moves from the point \((1, 1, 1)\) to the origin is:
\[ W = 5 \, \text{joules} \]