Problem 5 - Applied Physics

Question

The work required to lift a crate is \(31 \overline{0} \mathrm{~J}\). If the crate is lifted in \(25.0 \mathrm{~s}\), what power is developed?

Step-by-Step Solution

Answer

The power developed is 1,240 watts.

Step 1: Identify the formula for power

Power is defined as the rate at which work is done. Mathematically, power \( P \) can be calculated using the formula: \[ P = \frac{W}{t} \] where \( W \) is the work done in joules and \( t \) is the time in seconds.

Step 2: Substitute the given values

We are given the work \( W = 31,000 \) J (since \( 31 \overline{0} \) indicates 31,000 with one trailing zero) and the time \( t = 25 \) s. Substituting these values into the formula, we get: \[ P = \frac{31,000}{25} \]

Step 3: Perform the calculation

Divide \( 31,000 \) by \( 25 \) to find the power: \[ P = \frac{31,000}{25} = 1,240 \text{ W} \] This shows that the power developed is 1,240 watts.

Key Concepts

Power Calculation

Power in physics is a measure of how quickly work can be done over time. This means we can compute the power by dividing the total work done by the time it takes to perform that work. The formula for power is expressed as:\[ P = \frac{W}{t} \]Where:

• \( P \) is the power in watts (W)
• \( W \) represents the work done in joules (J)
• \( t \) is the time in seconds (s)

By understanding this simple formula, you can solve many physics problems involving power. In the given exercise, substituting the known values of work and time, which are 31,000 joules and 25 seconds respectively, we can calculate the power \[ P = \frac{31,000}{25} = 1,240 \text{ W} \]Thus, the power developed in lifting the crate is 1,240 watts.

Physics Problem-Solving

Physics problems, like the one in this exercise, are often solved by breaking the problem down into clear steps. Here's a simple approach:

• **Identify Given and Unknown Variables**: Clearly mark the values provided in the problem, for instance, work done or time taken.
• **Select Appropriate Formula**: Choose the formula that matches the structure of the problem. In our example, it's the formula for power.
• **Substitute the Values**: Place the known values into the formula, like we did with the joules and seconds.
• **Calculate and Solve**: Perform the calculation to find the unknown variable, such as power in this case.

Breaking down physics problems into manageable steps helps in avoiding errors and understanding each part of the problem-enhancing problem-solving skills.

Work-Energy Principle

The work-energy principle is a core concept in physics, stating that the work done on an object is equivalent to the change in its kinetic energy. This principle helps us understand how forces acting on an object affect its motion. ### Relevance to Power Calculation In the context of power, work is an integral factor because:

• The amount of work done relates directly to the energy transferred in a system.
• Understanding how work translates into energy changes allows us to grasp how efficient different processes are.
• Knowing the time over which this work occurs lets us compute power effectively.

This principle is fundamental when analyzing real-world systems. For instance, lifting a crate implies a force (like gravity) is overcome, converting energy into useful work. By mastering the work-energy principle, you can solve more complex problems involving different physical systems.