Problem 13 - Applied Physics

Question

A \(525-\mathrm{kg}\) steel beam is raised \(30.0 \mathrm{~m}\) in \(25.0 \mathrm{~s}\). How many kilowatts of power are needed?

Step-by-Step Solution

Answer

6.1659 kW

Step 1: Identify the Given Values

First, let us identify the information given in the exercise:- Mass of the steel beam, \( m = 525 \) kg- Height the beam is raised, \( h = 30.0 \) meters- Time taken to raise the beam, \( t = 25.0 \) seconds.

Step 2: Calculate the Work Done

The work done to raise the beam can be calculated using the formula for gravitational potential energy:\[ W = m \cdot g \cdot h \]where \( g = 9.81 \, \mathrm{m/s^2} \) (acceleration due to gravity). Substituting the given values:\[ W = 525 \, \mathrm{kg} \cdot 9.81 \, \mathrm{m/s^2} \cdot 30.0 \, \mathrm{m} \]\[ W = 154147.5 \, \mathrm{J} \]

Step 3: Calculate Power in Watts

Power is the rate at which work is done and is given by the formula:\[ P = \frac{W}{t} \]Substitute the calculated work and given time:\[ P = \frac{154147.5 \, \mathrm{J}}{25.0 \, \mathrm{s}} \]\[ P = 6165.9 \, \mathrm{W} \]

Step 4: Convert Power to Kilowatts

Finally, we need to convert the power from watts to kilowatts. Since 1 kilowatt (kW) equals 1000 watts, we use:\[ P = \frac{6165.9 \, \mathrm{W}}{1000} \]\[ P = 6.1659 \, \mathrm{kW} \]

Key Concepts

Gravitational Potential Energy

Understanding gravitational potential energy is crucial when analyzing situations involving objects being lifted vertically. Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It particularly depends on the height of the object relative to a reference point and the force of gravity. This concept is expressed mathematically as follows:\[ W = m \cdot g \cdot h \]where \( W \) is the work done or the gravitation potential energy in joules, \( m \) is the mass of the object in kilograms, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \mathrm{m/s^2} \)), and \( h \) is the height in meters.

• Gravitational potential energy highlights how much energy is stored due to the object's height.
• This energy doesn't depend on the path taken to reach the height, only on the initial and final positions.

The greater the height \( h \), or mass \( m \), of the object, the more gravitational potential energy it will have. This energy plays a pivotal role in solving physics problems like raising a steel beam, as in our exercise.

Work and Energy

Work and energy are fundamental concepts in physics, often used to describe similar ideas. Work in physics is defined as the energy transferred to an object via force over a distance. In simpler terms, it represents how much "effort" is used to move something. The concept of work can be expressed by the formula:\[ W = F \cdot d \cdot \cos(\theta) \]where \( W \) is work done measured in joules, \( F \) is the force applied in newtons, \( d \) is the displacement in meters, and \( \theta \) is the angle between the force and displacement direction. In the context of vertical lifting, angle \( \theta \) is zero, simplifying the equation to\[ W = F \cdot d \].

• Energy is the capacity to perform work.
• Work can change the energy of a system, such as increasing its kinetic or potential energy.

In our problem, since the displacement involves lifting, the force \( F \) is equal to the weight of the object, which is \( m \cdot g \). Understanding work in these terms allows comprehension of how potential energy changes during lifting processes like raising a steel beam.

Unit Conversion in Physics

Conversion of units is an essential aspect of solving physics problems, allowing us to express quantities in suitable units. A typical situation involves converting power from watts (W) to kilowatts (kW). Since 1 kilowatt is equal to 1000 watts, the conversion involves dividing the power in watts by 1000. This can be represented mathematically as follows:\[ P_{\text{kilowatts}} = \frac{P_{\text{watts}}}{1000} \]

• Unit conversion ensures consistent usage of units, particularly important in combined calculations like those involving work and energy.
• Always ensure units are compatible, particularly when inserting values into formulas.

In our example, the calculated power is initially found in watts, and converting it to kilowatts assists in obtaining a more practical and interpretable result. Consistent use of unit conversions can simplify and clarify problem analysis in physics.