To analyze and solve the problem in the uploaded image, I'll break it down step by step:

Problem Analysis

The graph provided in the image displays Force (N) versus Position (x in meters). Here's what we need to determine:

The question asks how the final velocity ($v_f$) of the cart at $x = 1 \, \text{m}$ compares to its initial velocity ($v_0$) at $x = 0 \, \text{m}$. Since the track is frictionless, the work-energy principle applies.

Key Concepts

1. Work-Energy Theorem: The work done by a force on an object is equal to the change in its kinetic energy: $W = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_0^2$ where $W$ is the work done, and $m$ is the mass of the cart.

2. Work Done from a Force vs. Position Graph: The work done is the area under the force-position graph between $x = 0$ and $x = 1$.

3. Force Analysis: Positive force increases velocity, while negative force decreases it.

Step-by-Step Solution

1. Calculate Work Done:

   • Between $x = 0$ and $x = 0.5 \, \text{m}$: Force is negative ($-1.0 \, \text{N}$), forming a triangular area. $W_1 = \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 0.5 \times (-1.0) = -0.25 \, \text{J}$
   • Between $x = 0.5 \, \text{m}$ and $x = 1.0 \, \text{m}$: Force is positive ($+1.0 \, \text{N}$), forming another triangular area. $W_2 = \text{Area} = \frac{1}{2} \times 0.5 \times (1.0) = 0.25 \, \text{J}$

2. Net Work Done: The total work done on the cart is: $W_{\text{total}} = W_1 + W_2 = -0.25 + 0.25 = 0 \, \text{J}$

3. Interpretation: Since the total work done on the cart is zero, the kinetic energy does not change. Therefore: $\Delta KE = 0 \implies \frac{1}{2} m v_f^2 = \frac{1}{2} m v_0^2 \implies v_f = v_0$

Conclusion

The final velocity ($v_f$) is equal to the initial velocity ($v_0$).

Correct Answer

$\boxed{v_f = v_0}$

Would you like further clarification or more details on the calculations?

Follow-up Questions

1. How can we calculate work from a graph with curved lines instead of straight lines?
2. How does friction affect the work-energy relationship in similar problems?
3. Can you explain the difference between positive and negative work in real-world scenarios?
4. What happens if the net work is not zero in this situation?
5. How would you calculate the final velocity if the mass of the cart were provided?

Tip

Always check the area under the curve carefully for both positive and negative contributions when analyzing force-position graphs!