Question

An engine works at \(30.0 \%\) efficiency. The engine raises a \(5.00-\mathrm{kg}\) crate from rest to a vertical height of \(10.0 \mathrm{m}\) at which point the crate has a speed of \(4.00 \mathrm{m} / \mathrm{s} .\) How much heat input is required for this engine?

Short answer

The required heat input is 743.3 J.

Step-by-step solution

Calculate work done against gravity

First, we calculate the work done to raise the crate to the height of 10.0 meters. This can be calculated using the formula for gravitational potential energy, \[ W_{gravity} = mgh \] where \( m = 5.00 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity), and \( h = 10.0 \text{ m} \). Calculating this gives us: \[ W_{gravity} = 5.00 \times 9.81 \times 10.0 = 490.5 \text{ J} \].

Key concepts

Efficiency

Efficiency is a measure of how well an engine converts input energy into useful work output. In simpler terms, it is about running a system using the minimum amount of energy to achieve the desired result. More specifically, efficiency (b7) can be expressed in terms of energy as:\[ b7 = \frac{W_{output}}{Q_{input}} \times 100\% \]Where:

• b7: Efficiency of the engine (unitless, often expressed as a percentage)
• \( W_{output} \): Useful work output (in Joules)
• \( Q_{input} \): Heat input or total energy input (in Joules)

For example, when an engine runs at a 30% efficiency, it means that only 30% of the total energy input is used for work, while the remaining 70% might be lost as waste heat or energy loss.Understanding efficiency is crucial because it dictates how much fuel or energy an engine consumes relative to the work it produces.

Work against gravity

When an object is elevated against the force of gravity, work is done. This work is linked directly to gravitational forces and increases with both the mass of the object and the height it is lifted.The formula for calculating the work done against gravity is:\[ W_{gravity} = mgh \]Where:

• \( m \): Mass of the object (in kg)
• \( g \): Acceleration due to gravity (approximately 9.81 m/s² on Earth)
• \( h \): Height to which the object is raised (in meters)

In the case of the exercise, the work against gravity was calculated to be 490.5 Joules. This signifies the energy needed to lift the crate without considering other factors like friction or air resistance.Work against gravity is an essential part of understanding energy transfer in mechanical systems.

Gravitational potential energy

Gravitational potential energy is the energy stored by an object due to its position relative to a reference point, usually near the Earth's surface. It is dependent on three primary factors:

• The mass of the object \( m \)
• The height \( h \) of the object above the chosen reference point
• The gravitational field strength \( g \)

The potential energy formula is:\[ PE_{gravity} = mgh \]This formula is fundamental when calculating the energy required to lift objects in gravitational fields and is used in various applications from roller coasters to space exploration. For instance, in raising a 5.00 kg crate to a height of 10.0 m, the potential energy is exactly that of the work against gravity: 490.5 Joules.

Heat input

Heat input is the total energy supplied to a system to perform work and overcome energy losses. In thermodynamic systems, this energy is generally supplied in the form of heat, hence the term "heat input."Given the efficiency equation:\[ b7 = \frac{W_{output}}{Q_{input}} \]We can rearrange it to solve for heat input:\[ Q_{input} = \frac{W_{output}}{b7} \]For an engine operating at 30% efficiency, the input required to perform 490.5 Joules of work can be calculated by inserting the efficiency and work output into the formula.Understanding the concept of heat input helps in grasping how much energy is consumed by an engine initially to produce a desired amount of work and to identify potential areas for improving energy consumption by increasing efficiency.