What is the work to be done to increase the velocity of a car from `30 km h^(-1)` to `60 km h^(-1)` if the mass of the car is `1500 kg` ?

Options

A. `156200J`
B. `156250J`
C. `156275J`
D. `156300J`

Explanation

To find the work done to increase the velocity of a car from \(30 \, \text{km/h}\) to \(60 \, \text{km/h}\) with a mass of \(1500 \, \text{kg}\), we can follow these steps:

### Step 1: Convert velocities from km/h to m/s
To convert the velocities from kilometers per hour to meters per second, we use the conversion factor \( \frac{5}{18} \).

- For \( V_1 = 30 \, \text{km/h} \):
\[ V_1 = 30 \times \frac{5}{18} = \frac{150}{18} = \frac{25}{3} \, \text{m/s} \]

- For \( V_2 = 60 \, \text{km/h} \):
\[ V_2 = 60 \times \frac{5}{18} = \frac{300}{18} = \frac{50}{3} \, \text{m/s} \]

### Step 2: Calculate the change in kinetic energy
The work done on the car is equal to the change in kinetic energy. The formula for kinetic energy (KE) is:
\[ KE = \frac{1}{2} m v^2 \]
Thus, the change in kinetic energy (\( \Delta KE \)) is given by:
\[ \Delta KE = KE_{final} - KE_{initial} = \frac{1}{2} m V_2^2 - \frac{1}{2} m V_1^2 \]

### Step 3: Substitute the values into the equation
Substituting the values for mass and velocities:
\[ \Delta KE = \frac{1}{2} \times 1500 \times \left( \left( \frac{50}{3} \right)^2 - \left( \frac{25}{3} \right)^2 \right) \]

### Step 4: Calculate \( V_2^2 \) and \( V_1^2 \)
Calculating the squares:
\[ V_2^2 = \left( \frac{50}{3} \right)^2 = \frac{2500}{9} \]
\[ V_1^2 = \left( \frac{25}{3} \right)^2 = \frac{625}{9} \]

### Step 5: Find the difference in kinetic energy
Now, substituting back into the equation:
\[ \Delta KE = \frac{1}{2} \times 1500 \times \left( \frac{2500}{9} - \frac{625}{9} \right) \]
\[ = \frac{1}{2} \times 1500 \times \left( \frac{2500 - 625}{9} \right) \]
\[ = \frac{1}{2} \times 1500 \times \left( \frac{1875}{9} \right) \]

### Step 6: Calculate the work done
Calculating the work done:
\[ \Delta KE = \frac{1500}{2} \times \frac{1875}{9} = 750 \times \frac{1875}{9} \]
\[ = \frac{1406250}{9} \approx 156250 \, \text{J} \]

### Final Answer
The work done to increase the velocity of the car from \(30 \, \text{km/h}\) to \(60 \, \text{km/h}\) is approximately:
\[ \boxed{156250 \, \text{J}} \]