To solve for the average power output ($P$) of the dragster, we use the formula:

$P = \frac{W_{\text{net}}}{t}$

Where:

• $W_{\text{net}}$ is the net work done on the dragster.
• $t = 9.7 \, \text{s}$ is the time taken.

Step 1: Net Work Done

The net work ($W_{\text{net}}$) is the difference between the total work done by the engine and the work done against friction. Total work is calculated as the kinetic energy change ($W_{\text{KE}}$) minus work done against friction ($W_{\text{friction}}$):

$W_{\text{net}} = W_{\text{KE}} - W_{\text{friction}}$

(a) Kinetic Energy Work ($W_{\text{KE}}$):

$W_{\text{KE}} = \frac{1}{2} m v^2$ Substitute $m = 515 \, \text{kg}$ and $v = 110 \, \text{m/s}$: $W_{\text{KE}} = \frac{1}{2} (515)(110^2)$ $W_{\text{KE}} = \frac{1}{2} (515)(12100) = 311,5750 \, \text{J}$

(b) Work Against Friction ($W_{\text{friction}}$):

$W_{\text{friction}} = f_{\text{friction}} \cdot d$ Substitute $f_{\text{friction}} = 1400 \, \text{N}$ and $d = 415 \, \text{m}$: $W_{\text{friction}} = 1400 \cdot 415 = 581,000 \, \text{J}$

(c) Net Work ($W_{\text{net}}$):

$W_{\text{net}} = W_{\text{KE}} - W_{\text{friction}} = 3,151,750 - 581,000 = 2,570,750 \, \text{J}$

Step 2: Average Power Output

$P = \frac{W_{\text{net}}}{t}$ Substitute $W_{\text{net}} = 2,570,750 \, \text{J}$ and $t = 9.7 \, \text{s}$: $P = \frac{2,570,750}{9.7} \approx 265,031.96 \, \text{W}$

Final Answer:

$P \approx 265,032 \, \text{W}$

Let me know if you'd like further clarification!

5 Related Questions

1. How is power related to energy and time in physical systems?
2. What factors influence the dragster's energy loss to friction?
3. What are the implications of increasing the dragster’s mass on its power output?
4. How does the work-energy principle simplify motion problems like this one?
5. What role does the time duration play in determining average power?

Tip: Always identify forces acting against motion, such as friction, to accurately calculate net work done.