A car loses 200 kJ of kinetic energy when its speed decreases from 25 m/s to 15 m/s. Find its mass in tonnes.
1
The question asks us to find the mass of a car given the change in its kinetic energy and its initial and final speeds. Kinetic energy is the energy an object possesses due to its motion. When the speed of an object changes, its kinetic energy also changes. The relationship between kinetic energy (KE), mass (m), and speed (v) is given by the formula:
\(\text{KE} = \frac{1}{2}mv^2\)
The change in kinetic energy (\(\Delta \text{KE}\)) is the difference between the final kinetic energy (\(\text{KE}_{final}\)) and the initial kinetic energy (\(\text{KE}_{initial}\)).
\(\Delta \text{KE} = \text{KE}_{final} - \text{KE}_{initial}\)
Since \(\text{KE} = \frac{1}{2}mv^2\), we can write the change in kinetic energy as:
\(\Delta \text{KE} = \frac{1}{2}mv_{final}^2 - \frac{1}{2}mv_{initial}^2\)
We can factor out \(\frac{1}{2}m\):
\(\Delta \text{KE} = \frac{1}{2}m(v_{final}^2 - v_{initial}^2)\)
We are given the following information about the car:
First, let's convert the change in kinetic energy from kilojoules (kJ) to joules (J), which is the standard unit for energy in SI units. \(1 \, \text{kJ} = 1000 \, \text{J}\).
\(\Delta \text{KE} = -200 \, \text{kJ} = -200 \times 1000 \, \text{J} = -200,000 \, \text{J}\)
Now we can substitute the given values into the formula for the change in kinetic energy:
\(\Delta \text{KE} = \frac{1}{2}m(v_{final}^2 - v_{initial}^2)\)
\(-200,000 \, \text{J} = \frac{1}{2}m((15 \, \text{m/s})^2 - (25 \, \text{m/s})^2)\)
Calculate the squares of the speeds:
\(15^2 = 15 \times 15 = 225\)
\(25^2 = 25 \times 25 = 625\)
Substitute these values back into the equation:
\(-200,000 = \frac{1}{2}m(225 - 625)\)
Calculate the difference in the squared speeds:
\(225 - 625 = -400\)
Substitute this difference back into the equation:
\(-200,000 = \frac{1}{2}m(-400)\)
Multiply \(\frac{1}{2}\) by \(-400\):
\(\frac{1}{2} \times -400 = -200\)
So the equation becomes:
\(-200,000 = -200m\)
Now, solve for the mass \(m\) by dividing both sides by \(-200\):
\(m = \frac{-200,000}{-200}\)
\(m = 1000 \, \text{kg}\)
The question asks for the mass in tonnes. The conversion factor between kilograms (kg) and tonnes is:
\(1 \, \text{tonne} = 1000 \, \text{kg}\)
To convert the mass from kilograms to tonnes, we divide the mass in kilograms by 1000:
Mass in tonnes = \(\frac{\text{Mass in kg}}{1000}\)
Mass in tonnes = \(\frac{1000 \, \text{kg}}{1000 \, \text{kg/tonne}}\)
Mass in tonnes = \(1 \, \text{tonne}\)
The mass of the car is 1000 kg, which is equal to 1 tonne. Therefore, the mass of the car in tonnes is 1.
| Step | Description | Calculation |
|---|---|---|
| 1 | Convert \(\Delta \text{KE}\) to Joules | \(-200 \, \text{kJ} = -200,000 \, \text{J}\) |
| 2 | Write formula for \(\Delta \text{KE}\) | \(\Delta \text{KE} = \frac{1}{2}m(v_{final}^2 - v_{initial}^2)\) |
| 3 | Substitute known values | \(-200,000 = \frac{1}{2}m((15)^2 - (25)^2)\) |
| 4 | Calculate squared speeds | \(15^2 = 225\), \(25^2 = 625\) |
| 5 | Substitute squared speeds | \(-200,000 = \frac{1}{2}m(225 - 625)\) |
| 6 | Calculate difference | \(225 - 625 = -400\) |
| 7 | Simplify equation | \(-200,000 = \frac{1}{2}m(-400) = -200m\) |
| 8 | Solve for mass (\(m\)) in kg | \(m = \frac{-200,000}{-200} = 1000 \, \text{kg}\) |
| 9 | Convert mass to tonnes | \(1000 \, \text{kg} \times \frac{1 \, \text{tonne}}{1000 \, \text{kg}} = 1 \, \text{tonne}\) |
| Concept | Definition/Formula | Units |
|---|---|---|
| Kinetic Energy (KE) | Energy of motion, \(\text{KE} = \frac{1}{2}mv^2\) | Joules (J) |
| Mass (\(m\)) | Measure of inertia | Kilograms (kg), Tonnes |
| Speed (\(v\)) | Magnitude of velocity | Meters per second (m/s) |
| Change in KE (\(\Delta \text{KE}\)) | \(\text{KE}_{final} - \text{KE}_{initial}\) | Joules (J) |
| Energy Loss | Indicated by negative \(\Delta \text{KE}\) when speed decreases | Joules (J) |
| Unit Conversion | 1 tonne = 1000 kg | - |
Problems involving changes in kinetic energy often relate to the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy (\(W_{net} = \Delta \text{KE}\)). In this specific car problem, the loss of kinetic energy could be due to work done by friction, air resistance, or braking forces. However, the question provides the change in KE directly, simplifying the calculation needed to find the mass.
It's important to pay attention to units throughout the calculation. Using SI units (Joules for energy, kilograms for mass, meters per second for speed) consistently makes the calculation correct. The final answer might require conversion to a different unit, like tonnes in this case, as specified by the question.
When dealing with squares of speeds, remember that \((v)^2\) is always positive, but \((v_{final}^2 - v_{initial}^2)\) can be negative if \(v_{final}\) is less than \(v_{initial}\), indicating a decrease in speed and a loss of kinetic energy.
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