Problem 80 - Master Resource Book in JEE Main Physics

Question

A \(500 \mathrm{~kg}\) car, moving with a velocity of \(36 \mathrm{kmh}^{-1}\) on a straight road unidirectionally, doubles its velocity in one minute. The power delivered by the engine for doubling the velocity is (a) \(750 \mathrm{~W}\) (b) \(1050 \mathrm{~W}\) (c) \(1150 \mathrm{~W}\) (d) \(1250 \mathrm{~W}\)

Step-by-Step Solution

Answer

(d) 1250 W.

Step 1: Convert Velocity to Meters per Second

The initial velocity of the car is given as \(36 \text{ km/h}\). First, convert this to meters per second by using the conversion factor \(1 \text{ km/h} = \frac{5}{18} \text{ m/s}\). Thus, the initial velocity \(v_i\) becomes \(36 \times \frac{5}{18} = 10 \text{ m/s}\).

Step 2: Determine the Final Velocity

The car's velocity doubles. Therefore, the final velocity \(v_f\) is twice the initial velocity. Compute \(v_f = 2 \times 10 \text{ m/s} = 20 \text{ m/s}\).

Step 3: Calculate the Initial Kinetic Energy

The initial kinetic energy (\(KE_i\)) is given by the equation \(KE_i = \frac{1}{2} m v_i^2\). Substituting the values gives \(KE_i = \frac{1}{2} \times 500 \times 10^2 = 25000 \text{ J}\).

Step 4: Calculate the Final Kinetic Energy

The final kinetic energy (\(KE_f\)) is given by \(KE_f = \frac{1}{2} m v_f^2\). Substituting the values gives \(KE_f = \frac{1}{2} \times 500 \times 20^2 = 100000 \text{ J}\).

Step 5: Determine the Change in Kinetic Energy

The change in kinetic energy (\(\Delta KE\)) is the difference between the final and initial kinetic energies, \(\Delta KE = KE_f - KE_i = 100000 - 25000 = 75000 \text{ J}\).

Step 6: Calculate the Power Delivered by the Engine

Power is defined as work done over time. Here, the work done is equal to the change in kinetic energy, and the time is given as one minute (\(60 \text{ seconds}\)). Thus, the power \(P\) is \(P = \frac{75000}{60} = 1250 \text{ W}\).

Step 7: Choose the Correct Answer

The calculated power delivered by the engine is \(1250 \text{ W}\). So, the correct choice is (d) \(1250 \text{ W}\).

Key Concepts

Kinetic energy

Kinetic energy is the energy that an object possesses due to its motion. It is a fundamental concept in physics and can be calculated using the formula:\[ KE = \frac{1}{2} m v^2 \]where:

• \( KE \) is the kinetic energy,
• \( m \) is the mass of the object in kilograms,
• \( v \) is the velocity of the object in meters per second.

Understanding kinetic energy is crucial because it allows us to quantify how the velocity and mass of an object contribute to its overall energy. For example, in the given exercise, the car has a mass of 500 kg and its kinetic energy changes as its velocity increases. By calculating both initial and final kinetic energies, we can determine how much energy is needed to increase its speed, making this concept a key part of the power calculation.

Velocity conversion

Velocity conversion is an essential step when dealing with kinetic energy calculations, especially if the velocity provided is not in the standard scientific unit of meters per second (m/s). Many real-world problems require us to convert units to apply the formulas correctly.For the given exercise, the car's initial velocity is provided as \(36 \text{ km/h}\). To convert this to m/s, we multiply by the conversion factor \(\frac{5}{18}\). The formula used is:\[ v (\text{m/s}) = v (\text{km/h}) \times \frac{5}{18} \]This conversion helps ensure that all our calculations are consistent and accurate. After conversion, the velocity was found to be 10 m/s, which aligns with standard units for calculating kinetic energy.

Work-energy principle

The work-energy principle is fundamental in connecting the concepts of force, work, and energy. It states that the work done on an object is equal to the change in its kinetic energy. In simpler terms, the more work we do on something, the more its energy changes.In the given scenario, we first calculated the change in kinetic energy of the car as it increased its velocity. This change was found to be \(75000 \text{ J}\). According to the work-energy principle, this change in kinetic energy corresponds to the work done by the engine.Finally, to find the power, which is the rate of doing work, we use the formula:\[ P = \frac{\Delta KE}{\Delta t} \]where:

• \( P \) is the power,
• \( \Delta KE \) is the change in kinetic energy,
• \( \Delta t \) is the time interval.

So, the power delivered by the engine to double the velocity in one minute was calculated as 1250 W. This ties together the kinetic energy, velocity conversion, and work-energy principle concepts neatly for solving the problem.