The engine of a beach buggy develops 40 kW at a speed of 900 r/minutes.

Here's the solution for Question 2.2.

Given Data:

• Power ($P$) = 40 kW
• Rotational speed ($N$) = 900 r/minutes

Step 1: Convert power to Watts (W). $P = 40 kW = 40 \times 1000 W = 40000 \text{ W}$

Step 2: Convert rotational speed from revolutions per minute (r/minutes) to revolutions per second (r/s). $N_{rev/s} = \frac{900 r/minutes}{60 s/minute} = 15 \text{ r/s}$

Step 3: Calculate the angular velocity ($\omega$) in radians per second (rad/s). The angular velocity is given by $\omega = 2 \pi N_{rev/s}$. This is the value for 's' in the context of angular speed. $\omega = 2 \pi \times 15 \text{ r/s}$ $\omega = 30 \pi \text{ rad/s}$ $\omega \approx 30 \times 3.14159 \text{ rad/s}$ $\omega \approx \text{94.25 rad/s}$

Step 4: Calculate the Torque ($T$). The relationship between power, torque, and angular velocity is $P = T \omega$. $T = \frac{P}{\omega}$ $T = \frac{40000 W}{30 \pi rad/s}$ $T = \frac{40000}{30 \pi} N\cdot\text{m}$ $T = \frac{4000}{3 \pi} N\cdot\text{m}$ $T \approx \frac{4000}{3 \times 3.14159} N\cdot\text{m}$ $T \approx \frac{4000}{9.42477} N\cdot\text{m}$ $T \approx \text{424.41 N\cdotm}$

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