Question

A rope-way trolley of mass \(1200 \mathrm{~kg}\) uniformly from rest to a velocity of \(72 \mathrm{~km} / \mathrm{h}\) in \(6 \mathrm{~s}\). What is the average power of the engine during this period in watt ? (Neglect friction) (A) \(400 \mathrm{~W}\) (B) \(40,000 \mathrm{~W}\) (C) \(24000 \mathrm{~W}\) (D) \(4000 \mathrm{~W}\)

Short answer

The average power of the engine during this period is 40,000 W. The correct answer is (B) \(40,000 \mathrm{~W}\).

Step-by-step solution

Convert the given velocity from km/h to m/s

To convert the velocity from km/h to m/s, we use the conversion factor 1000/3600 (1 km = 1000 m, 1 hour = 3600 seconds): \[v = 72 \frac{\text{km}}{\text{h}} \cdot \frac{1000 \text{ m}}{1 \text{ km}} \cdot \frac{1 \text{ h}}{3600 \text{ s}} = 20 \frac{\text{m}}{\text{s}}\]

Calculate the final kinetic energy (K.E.) of the trolley

Now, we need to calculate the final kinetic energy of the trolley. The formula for kinetic energy is: \[ K.E. = \frac{1}{2}mv^2\] Using the given mass (1200 kg) and calculated velocity (20 m/s), we get: \[ K.E. = \frac{1}{2}\times 1200 \times (20)^2 = 240,000 \mathrm{~J}\]

Calculate the work done (W) on the trolley

The trolley started from rest so its initial kinetic energy was zero. The work done on the trolley is the difference between its final and initial kinetic energies. In this case, the initial K.E. is zero so the difference is the same as the final K.E.: \[W = K.E. - 0 = 240,000 \mathrm{~J}\]

Calculate the average power (P) of the engine

The average power (P) of the engine can be calculated using the formula: \[P = \frac{W}{t}\] where t is the time taken. We know t=6s and W=240,000J. \[P = \frac{240,000 \mathrm{~J}}{6 \mathrm{~s}} = 40,000 \mathrm{~W}\] The average power of the engine during this period is 40,000 W. The correct answer is (B) \(40,000 \mathrm{~W}\).

Key concepts

Kinetic Energy

When we consider kinetic energy, we think of the energy that an object possesses due to its motion. For any moving object, this energy can be calculated using the formula:

• \( K.E. = \frac{1}{2} mv^2 \)

Where \(m\) is the mass of the object, and \(v\) is its velocity. In our exercise, the mass of the rope-way trolley is 1200 kg and, after conversion, it moves with a velocity of 20 m/s.
Thus, by plugging the numbers into the kinetic energy formula, we get a kinetic energy of 240,000 J (Joules).
By understanding how to calculate kinetic energy, you'll see how velocity and mass directly contribute to the energy of a moving object. This fundamental concept helps bridge to more complex physics topics involving energy transfer and mechanics.

Conversion of Units

Let's tackle the important task of unit conversion. In physics, quantities often come in different units, and converting them correctly is crucial to solving problems accurately.
Here, we need to convert velocity from kilometers per hour (km/h) to meters per second (m/s). This is done using the conversion factor:

• 1000 meters = 1 kilometer
• 3600 seconds = 1 hour

To convert 72 km/h into m/s, the calculation is as follows:

• \(v = 72 \frac{\text{km}}{\text{h}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 20 \frac{\text{m}}{\text{s}}\)

This conversion ensures that all units in our equation are compatible, allowing us to proceed with further calculations using SI units, which are the standard in scientific studies.

Work Done

Work done is a measure of energy transfer when a force moves an object over a distance. It is calculated as the change in kinetic energy for objects with varying speeds.
In our scenario, the trolley starts from rest. This implies its initial kinetic energy is zero. Once it reaches a speed of 20 m/s, its final kinetic energy is 240,000 J. The work done on the trolley, therefore, is the change in kinetic energy. Since it starts at rest, the initial work done is zero:

• \(W = \text{Final K.E.} - 0 = 240,000 \text{ J}\)

Understanding work involves applying the concept of energy change as forces are applied, fundamental to various applications in mechanics and engineering.

Physics Problem Solving

Solving physics problems often requires a strategic approach. Let's break this down using our exercise as an example. 1. **Identifying Known Quantities**: We start by examining what is given: the mass (1200 kg), initial and final velocities (0 to 72 km/h), and the time taken (6 seconds).2. **Correct Unit Conversion**: Convert units to ensure consistency and accuracy in calculations. This converts 72 km/h into 20 m/s.3. **Apply Newtonian Principles**: Use formulas like kinetic energy \((K.E. = \frac{1}{2}mv^2)\) and work-energy principles \((W = ext{Final K.E.} - ext{Initial K.E.})\) to find work done.4. **Calculate Average Power**: Use the power formula \((P = \frac{W}{t})\) to find the average power. Substituting work and time gives us a power of 40,000 W.
Through systematic steps, you gain proficiency in translating a real-world situation into manageable physics terms, paving the way for more advanced problem-solving skills.