7.5 Q-10

Question Statement

A force of magnitude 6 units is acting parallel to the vector $2 \hat{i} - 2 \hat{j} + \hat{k}$ and displaces the point of application from $(1, 2, 3)$ to $(5, 3, 7)$. Find the work done.

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Background and Explanation

To calculate the work done by a force, we use the formula:

$$W = \mathbf{F} \cdot \mathbf{d}$$

Where:

• $\mathbf{F}$ is the force vector,
• $\mathbf{d}$ is the displacement vector.

In this case, the force is acting in the direction of a vector, so we first need to calculate the unit vector in the direction of the given vector, and then scale it by the magnitude of the force. The displacement is calculated by subtracting the initial position from the final position.

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Solution

Step 1: Find the unit vector in the direction of the force

The force is acting in the direction of the vector $\mathbf{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}$. To find the unit vector, we first calculate the magnitude of $\mathbf{a}$:

$$|\mathbf{a}| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$

The unit vector $\mathbf{U}$ in the direction of $\mathbf{a}$ is given by:

$$\mathbf{U} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{2 \hat{i} - 2 \hat{j} + \hat{k}}{3}$$

Thus:

$$\mathbf{U} = \frac{2 \hat{i} - 2 \hat{j} + \hat{k}}{3}$$

Step 2: Calculate the force vector

The magnitude of the force is given as 6 units. The force vector $\mathbf{F}$ is:

$$\mathbf{F} = 6 \mathbf{U} = 6 \times \frac{2 \hat{i} - 2 \hat{j} + \hat{k}}{3}$$

Simplifying:

$$\mathbf{F} = 2 \left( 2 \hat{i} - 2 \hat{j} + \hat{k} \right) = 4 \hat{i} - 4 \hat{j} + 2 \hat{k}$$

Thus, the force vector is:

$$\mathbf{F} = 4 \hat{i} - 4 \hat{j} + 2 \hat{k}$$

Step 3: Find the displacement vector

The displacement vector $\mathbf{d}$ is calculated by subtracting the initial position $(1, 2, 3)$ from the final position $(5, 3, 7)$:

$$\mathbf{d} = (5 - 1) \hat{i} + (3 - 2) \hat{j} + (7 - 3) \hat{k}$$ $$\mathbf{d} = 4 \hat{i} + \hat{j} + 4 \hat{k}$$

Thus, the displacement vector is:

$$\mathbf{d} = 4 \hat{i} + \hat{j} + 4 \hat{k}$$

Step 4: Calculate the work done

The work done is the dot product of the force vector $\mathbf{F}$ and the displacement vector $\mathbf{d}$:

$$W = \mathbf{F} \cdot \mathbf{d}$$

Substitute the components of the vectors:

$$W = (4 \hat{i} - 4 \hat{j} + 2 \hat{k}) \cdot (4 \hat{i} + \hat{j} + 4 \hat{k})$$

Now, compute the dot product:

$$W = 4(4) + (-4)(1) + 2(4)$$ $$W = 16 - 4 + 8 = 20$$

Thus, the work done is:

$$W = 20 , \text{units}$$

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Key Formulas or Methods Used

• Work Done by a Force:

$$W = \mathbf{F} \cdot \mathbf{d}$$

Where $\mathbf{F}$ is the force vector and $\mathbf{d}$ is the displacement vector.

• Unit Vector:

$$\mathbf{U} = \frac{\mathbf{a}}{|\mathbf{a}|}$$

The unit vector is the vector $\mathbf{a}$ normalized by dividing it by its magnitude.

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Summary of Steps

1. Calculate the unit vector: Find the unit vector in the direction of the given vector $\mathbf{a}$ by dividing it by its magnitude.
2. Determine the force vector: Multiply the unit vector by the magnitude of the force to get the force vector.
3. Calculate the displacement vector: Subtract the initial position from the final position to find the displacement vector.
4. Compute the work done: Take the dot product of the force vector and the displacement vector to find the work done, which is $20$ units.