To solve this problem, we will use the provided formula for the final velocity $v_f$:

$v_f = \sqrt{\frac{mv_i^2 + 2F d \cos(\theta)}{m}}$

Step 1: Identify the variables

• Mass of the go-kart: $m = 125 \, \text{kg}$
• Initial velocity: $v_i = 8.75 \, \text{m/s}$
• Force applied: $F = 436 \, \text{N}$
• Distance moved: $d = 9.03 \, \text{m}$
• Angle between force and displacement: $\cos(\theta) = \cos(0^\circ) = 1$ (since force is along the direction of motion)

Step 2: Simplify the formula

We simplify the numerator: $mv_i^2 + 2F d \cos(\theta) = (125)(8.75)^2 + 2(436)(9.03)(1)$

Step 3: Perform the calculations

1. Compute $mv_i^2$: $mv_i^2 = 125 \cdot (8.75)^2 = 125 \cdot 76.5625 = 9570.31 \, \text{J}$

2. Compute $2Fd\cos(\theta)$: $2Fd\cos(\theta) = 2 \cdot 436 \cdot 9.03 = 7870.32 \, \text{J}$

3. Add the terms: $mv_i^2 + 2Fd\cos(\theta) = 9570.31 + 7870.32 = 17440.63 \, \text{J}$

4. Divide by the mass $m$: $\frac{mv_i^2 + 2Fd\cos(\theta)}{m} = \frac{17440.63}{125} = 139.525 \, \text{m}^2/\text{s}^2$

5. Take the square root to find $v_f$: $v_f = \sqrt{139.525} \approx 11.81 \, \text{m/s}$

Final Answer:

The final velocity of the go-kart is approximately: $v_f = +11.81 \, \text{m/s}$

This velocity is positive because the motion is in the forward direction.

Do you want further explanation or clarification? 😊

Here are 5 follow-up questions:

1. How would the final velocity change if the initial velocity were zero?
2. What would happen to the final velocity if the applied force were halved?
3. Why do we assume $\cos(\theta) = 1$ in this scenario?
4. Can this formula be used for vertical motion? Why or why not?
5. What if friction were present? How would that alter the calculations?

Tip: Always double-check the angle in physics problems involving force to ensure you're correctly applying $\cos(\theta)$.