Problem 68 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A \(1400-\mathrm{kg}\) sports car accelerates from rest to \(95 \mathrm{~km} / \mathrm{h}\) in \(7.4 \mathrm{~s}\). What is the average power delivered by the engine?

Step-by-Step Solution

Answer

The average power delivered by the engine is approximately 65.82 kW.

Step 1: Convert Velocity to Meters per Second

The final velocity is given as \(95 \text{ km/h}\). First, we need to convert this velocity to meters per second (m/s) using the conversion: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). Thus, \(95 \text{ km/h} = \frac{95}{3.6} \text{ m/s} \approx 26.39 \text{ m/s}\).

Step 2: Calculate the Kinetic Energy

The car accelerates from rest, so its initial velocity is \(0 \text{ m/s}\). The kinetic energy at the final velocity is given by: \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity. Substituting the given values: \(KE = \frac{1}{2} \times 1400 \times (26.39)^2\). Calculate to get \(KE \approx 486,869 \text{ J}\).

Step 3: Calculate the Work Done

Since the car accelerates to this speed from rest, the work done on the car is equal to its change in kinetic energy. With initial kinetic energy being zero (since initial speed is zero), the work done is \(486,869 \text{ J}\).

Step 4: Calculate the Average Power

Average power is the work done divided by the time period over which the work was done. The formula for power is \(P = \frac{W}{t}\) where \(W\) is work and \(t\) is time. Here, the work \(W = 486,869 \text{ J}\) and \(t = 7.4 \text{ s}\). Hence, \(P = \frac{486,869}{7.4} \approx 65,818 \text{ W} \approx 65.82 \text{ kW}\).

Key Concepts

Kinetic Energy

Kinetic energy, in simple terms, is the energy an object possesses due to its motion. It’s one of the most fundamental principles in physics, especially when it comes to dynamics and mechanical systems. Anytime an object moves, it has kinetic energy. The formula to calculate this energy is: \[ KE = \frac{1}{2}mv^2 \] where:

• \( m \) is the mass of the object, and
• \( v \) is the velocity of the object.

In our exercise, a sports car with a mass of 1400 kg increases its speed to 95 km/h. To find its kinetic energy in motion, we use the mass and its final velocity in meters per second. Remember, when an object starts from rest, all kinetic energy is gained through acceleration.

Work-Energy Theorem

The work-energy theorem is an important principle that bridges the concepts of work and kinetic energy. It states that the work done by forces on an object results in a change in its kinetic energy. So essentially, the work done on an object is equal to the change in its kinetic energy. This can be mathematically expressed as: \[ W = \Delta KE \] For our sports car scenario, the work done by its engine is the amount of energy transferred as the car accelerates to its final speed. Since it starts from rest, all the work goes into achieving its final kinetic energy, which is calculated using its mass and velocity. This relationship helps us understand energy transfer, such as how engines use fuel energy to increase a vehicle’s speed.

Unit Conversion

Unit conversion is a crucial step in solving many physics problems, particularly when dealing with SI units. In physics, it's important to use a consistent unit system to ensure accurate calculations. For velocity, converting from kilometers per hour (km/h) to meters per second (m/s) is essential because the standard unit of speed in the International System (SI) is m/s. The conversion factor from km/h to m/s is: \[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]In our detailed solution, the velocity of the car was converted from 95 km/h to approximately 26.39 m/s. This conversion ensures that when we plug the velocity into formulas, like kinetic energy, we are using compatible units. Always check the units, and make sure they align with the equations being used in any physics calculation.