The mass of cyclist together with the bike is 90 kg. Calculate the increse in kinetic energy if the speed increses from 6.0 km/h to 12 km/h.

Explanation

To solve the problem of calculating the increase in kinetic energy when the speed of a cyclist (along with the bike) increases from 6.0 km/h to 12.0 km/h, we will follow these steps:

### Step-by-Step Solution:

1. **Convert Speeds from km/h to m/s**:
- The formula to convert km/h to m/s is:
\[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18} \]
- For the initial speed (6.0 km/h):
\[ u = 6.0 \times \frac{5}{18} = \frac{30}{18} = \frac{5}{3} \text{ m/s} \]
- For the final speed (12.0 km/h):
\[ v = 12.0 \times \frac{5}{18} = \frac{60}{18} = \frac{10}{3} \text{ m/s} \]

2. **Calculate Initial Kinetic Energy (KE_initial)**:
- The formula for kinetic energy is:
\[ KE = \frac{1}{2} mv^2 \]
- Using the mass \( m = 90 \) kg and initial speed \( u = \frac{5}{3} \) m/s:
\[ KE_{\text{initial}} = \frac{1}{2} \times 90 \times \left(\frac{5}{3}\right)^2 \]
\[ = \frac{1}{2} \times 90 \times \frac{25}{9} = 45 \times \frac{25}{9} = \frac{1125}{9} \text{ J} \]

3. **Calculate Final Kinetic Energy (KE_final)**:
- Using the final speed \( v = \frac{10}{3} \) m/s:
\[ KE_{\text{final}} = \frac{1}{2} \times 90 \times \left(\frac{10}{3}\right)^2 \]
\[ = \frac{1}{2} \times 90 \times \frac{100}{9} = 45 \times \frac{100}{9} = \frac{4500}{9} \text{ J} \]

4. **Calculate the Increase in Kinetic Energy (ΔKE)**:
- The increase in kinetic energy is given by:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
\[ = \frac{4500}{9} - \frac{1125}{9} = \frac{4500 - 1125}{9} = \frac{3375}{9} \text{ J} \]
\[ = 375 \text{ J} \]

### Final Answer:
The increase in kinetic energy is **375 J**.