A 1.5 kg cart moves along the x axis at 2.0 m/s when it passes x = 1 m…

Step 1. Work. $W = \Big[\tfrac23 x^3 + 3x\Big]_1^4 = \left(\tfrac{128}{3} + 12\right) - \left(\tfrac23 + 3\right) = 51.0\ \text{J}$.

Step 2. Energy at the start. $\tfrac12 \times 1.5\ \text{kg} \times (2.0\ \text{m/s})^2 = 3.0\ \text{J}$.

Step 3. Speed. $\tfrac12 \times 1.5\ \text{kg} \times v^2 = 3.0\ \text{J} + 51.0\ \text{J} = 54\ \text{J}$, so $v^2 = 72\ \text{m}^2/\text{s}^2$ and $v = 8.49\ \text{m/s}$.