This problem involves calculating the work done by force vectors $\vec{F}$ in the direction of $\overrightarrow{PQ}$, for three different forces. The formula for work $W$ done by a force in the direction of a displacement is:

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

Where:

• $\vec{F}$ is the force vector.
• $\vec{d}$ is the displacement vector, which is $\overrightarrow{PQ}$ in this case.
• $\cdot$ represents the dot product of the two vectors.

The vector $\overrightarrow{PQ} = \langle 2, 0 \rangle$.

We need to find the dot product of $\vec{F}$ with $\overrightarrow{PQ}$ in each case:

(a) $\vec{F} = \langle 1, -3 \rangle$

$W_a = \vec{F} \cdot \overrightarrow{PQ} = \langle 1, -3 \rangle \cdot \langle 2, 0 \rangle = (1)(2) + (-3)(0) = 2$ The work done in case (a) is $2$ units.

(b) $\vec{F} = \langle 0, 3 \rangle$

$W_b = \vec{F} \cdot \overrightarrow{PQ} = \langle 0, 3 \rangle \cdot \langle 2, 0 \rangle = (0)(2) + (3)(0) = 0$ The work done in case (b) is $0$ units.

(c) $\vec{F} = \langle -3, 0 \rangle$

$W_c = \vec{F} \cdot \overrightarrow{PQ} = \langle -3, 0 \rangle \cdot \langle 2, 0 \rangle = (-3)(2) + (0)(0) = -6$ The work done in case (c) is $-6$ units.

Summary of Results:

• (a) Work = 2 units
• (b) Work = 0 units
• (c) Work = -6 units

If you'd like sketches for each case or further explanation on any of these calculations, let me know!

Do you have any questions or would you like further details?

Here are five related questions to deepen your understanding:

1. How is the direction of the displacement vector important in calculating work?
2. What does a negative work value signify physically?
3. How would the work change if the displacement vector had different components?
4. What is the significance of the dot product in physics for work?
5. How does the angle between force and displacement affect the amount of work done?

Tip: If the force is perpendicular to the displacement, the work done will always be zero, as shown in case (b).