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

Explanation

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

### Step 1: Convert speeds from km/h to m/s
First, we need to convert the speeds from kilometers per hour (km/h) to meters per second (m/s) using the conversion factor \(1 \text{ km/h} = \frac{5}{18} \text{ m/s}\).

- Initial speed \(v_i = 6.0 \text{ km/h} = 6.0 \times \frac{5}{18} = \frac{30}{18} = \frac{5}{3} \text{ m/s}\)
- Final speed \(v_f = 12.0 \text{ km/h} = 12.0 \times \frac{5}{18} = \frac{60}{18} = \frac{10}{3} \text{ m/s}\)

### Step 2: Use the kinetic energy formula
The kinetic energy (KE) is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \(m\) is the mass and \(v\) is the velocity.

### Step 3: Calculate initial kinetic energy
Using the initial speed \(v_i = \frac{5}{3} \text{ m/s}\):
\[ KE_i = \frac{1}{2} \times 90 \times \left(\frac{5}{3}\right)^2 \]
Calculating:
\[ KE_i = \frac{1}{2} \times 90 \times \frac{25}{9} = 45 \times \frac{25}{9} = \frac{1125}{9} \text{ J} \]

### Step 4: Calculate final kinetic energy
Using the final speed \(v_f = \frac{10}{3} \text{ m/s}\):
\[ KE_f = \frac{1}{2} \times 90 \times \left(\frac{10}{3}\right)^2 \]
Calculating:
\[ KE_f = \frac{1}{2} \times 90 \times \frac{100}{9} = 45 \times \frac{100}{9} = \frac{4500}{9} \text{ J} \]

### Step 5: Calculate the increase in kinetic energy
The increase in kinetic energy is given by:
\[ \Delta KE = KE_f - KE_i \]
Calculating:
\[ \Delta KE = \frac{4500}{9} - \frac{1125}{9} = \frac{4500 - 1125}{9} = \frac{3375}{9} = 375 \text{ J} \]

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

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