To solve this problem, we will use the Work-Energy Theorem, which states:

$W = \Delta KE$

where:

• $W$ is the work done by the force,
• $\Delta KE$ is the change in kinetic energy,
• $KE = \frac{1}{2} m v^2$ is the kinetic energy.

Step 1: Find the Work Done by the Force

Work is calculated as the area under the force-position graph from $x = 4$ m to $x = 6$ m.

From the graph:

• Between $x = 4$ m and $x = 5$ m, the force is 20 N (rectangle of area $20 \times 1 = 20$).

• Between $x = 5$ m and $x = 6$ m, the force decreases linearly from 20 N to 0 N, forming a right triangle with base $1$ and height $20$, so its area is:

  $\frac{1}{2} \times 1 \times 20 = 10$

Thus, total work done from $x = 4$ m to $x = 6$ m:

$W = 20 + 10 = 30 \text{ J}$

Step 2: Apply the Work-Energy Theorem

Since $m = 1$ kg, the change in kinetic energy is:

$30 = \frac{1}{2} (1) v^2 - \frac{1}{2} (1) (9)^2$

$30 = \frac{1}{2} v^2 - \frac{1}{2} (81)$

$30 = \frac{1}{2} v^2 - 40.5$

$30 + 40.5 = \frac{1}{2} v^2$

$70.5 = \frac{1}{2} v^2$

$v^2 = 141$

$v = \sqrt{141} \approx 11.87 \text{ m/s}$

Final Answer:

$\mathbf{11.87 \text{ m/s}}$

Would you like a more detailed explanation or have any questions?

Related Questions:

1. How is work calculated from a force-position graph?
2. What is the significance of the Work-Energy Theorem?
3. How would the answer change if the object had a different mass?
4. What happens if the force were negative instead of positive?
5. How would you determine velocity at any other position (not just $x = 6$)?

Tip:

Always check the force graph carefully for distinct regions and use area calculations accordingly! 🚀