The work done by a force `vecF` during a displacement `vecr` is given by `vecF.vecr`. Suppose a force of 12 N acts on a particle in vertically upward directionand the particle is displaced through 2.0 m in vertically downward direction. Find the work done by the force during this displacement.

Explanation

To find the work done by the force during the displacement, we can follow these steps:

### Step 1: Identify the Force and Displacement Vectors
- The force \( \vec{F} \) is acting in the vertically upward direction with a magnitude of 12 N. We can represent this as:
\[ \vec{F} = 12 \hat{k} \]
- The displacement \( \vec{r} \) is in the vertically downward direction with a magnitude of 2 m. We can represent this as:
\[ \vec{r} = -2 \hat{k} \]

### Step 2: Use the Work Done Formula
The work done \( W \) by a force during a displacement is given by the dot product of the force and displacement vectors:
\[ W = \vec{F} \cdot \vec{r} \]

### Step 3: Substitute the Vectors into the Formula
Now we can substitute the vectors into the work done formula:
\[ W = (12 \hat{k}) \cdot (-2 \hat{k}) \]

### Step 4: Calculate the Dot Product
The dot product of two vectors can be calculated as follows:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta) \]
In this case, since both vectors are in the same direction (along the \( \hat{k} \) axis), the angle \( \theta \) between them is 180 degrees. Thus, \( \cos(180^\circ) = -1 \):
\[ W = 12 \times (-2) \times (\hat{k} \cdot \hat{k}) = 12 \times (-2) \times 1 = -24 \text{ Joules} \]

### Step 5: Conclusion
The work done by the force during this displacement is:
\[ W = -24 \text{ Joules} \]

This negative sign indicates that the force and displacement are in opposite directions, which is consistent with the physical situation described.