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Chapter 1: Problem 10

An object weighs \(25 \mathrm{kN}\) at a location where the acceleration of gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\). Determine its mass, in \(\mathrm{kg}\).

Short Answer

Expert verified
The mass is approximately 2551.02 kg.

Step by step solution

01

Understand the relationship between weight and mass

The weight of an object is given by the formula: \[ W = m \times g \] where:- \(W\) is the weight,- \(m\) is the mass, and- \(g\) is the acceleration due to gravity.
02

Identify the given values

From the problem statement, the given values are:- Weight, \(W = 25 \text{kN} = 25000 \text{N} \)- Acceleration due to gravity, \( g = 9.8 \text{ m/s}^2 \)
03

Rearrange the formula to solve for mass

Rearrange the weight formula to solve for mass, \( m \): \[ m = \frac{W}{g} \]
04

Substitute the given values into the rearranged formula

Substitute \( W = 25000 \text{N} \) and \( g = 9.8 \text{m/s}^2 \) into the formula: \[ m = \frac{25000 \text{N}}{9.8 \text{m/s}^2} \]
05

Perform the calculation

Calculate the mass: \[ m = \frac{25000}{9.8} \approx 2551.02 \text{kg} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Weight and Mass Relationship
Weight and mass are commonly confused, but they are distinct concepts. Mass is the amount of matter in an object and is measured in kilograms (kg).
Weight, on the other hand, is the force exerted by gravity on that mass and is measured in newtons (N).
The relationship between weight and mass is given by the formula: \[ W = m \times g \] where
  • \(W\) is the weight,
  • \(m\) is the mass, and
  • \(g\) is the acceleration due to gravity.
This formula shows that weight depends on both mass and gravity.
For instance, the same object will weigh less on the Moon than on Earth because the Moon's gravitational pull is weaker.
Acceleration Due to Gravity
Acceleration due to gravity (\(g\)) is the rate at which objects accelerate downwards due to Earth's gravitational pull. On Earth, this value is approximately \(9.8 \text{ m/s}^2\).
It's an essential constant in the weight equation and can vary slightly depending on location (e.g., altitude and latitude).
Understanding \(g\) allows us to determine how massive objects are by measuring their weight. Therefore, if you know an object's weight and the local gravity, you can calculate its mass using the formula we derived earlier: \[ m = \frac{W}{g} \]
Unit Conversion in Physics
Unit conversion is crucial in physics to ensure that calculations are accurate and consistent. In this exercise, we converted the given weight from kilonewtons (kN) to newtons (N).
1 kN equals 1000 N, which simplifies the calculation.
That鈥檚 why the initial weight of \(25 \text{kN}\) became \(25000 \text{N}\).
Another example is length conversions, such as 1 kilometer (km) equals 1000 meters (m).
Always ensure consistent units before performing calculations to avoid errors.

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Most popular questions from this chapter

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A dish of liquid water is placed on a table in a room. After a while, all of the water evaporates. Taking the water and the air in the room to be a closed system, can the system be regarded as a pure substance during the process? After the process is completed? Discuss.

Over a limited temperature range, the relation between electrical resistance \(R\) and temperature \(T\) for a resistance temperature detector is $$ R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right] $$ where \(R_{0}\) is the resistance, in ohms \((\Omega)\), measured at reference temperature \(T_{0}\) (in \(\left.{ }^{\circ} \mathrm{C}\right)\) and \(\alpha\) is a material constant with units of \(\left({ }^{\circ} \mathrm{C}\right)^{-1}\). The following data are obtained for a particular resistance thermometer: \begin{tabular}{lcrr} \hline & \(T\left({ }^{\circ} \mathrm{C}\right)\) & \multicolumn{2}{c}{\(R(\Omega)\)} \\ \hline Test \(1\left(\mathrm{~T}_{0}\right)\) & 0 & \(\left(\mathrm{R}_{0}\right)\) & \(51.39\) \\ Test 2 & 91 & \(51.72\) \\ \hline \end{tabular} What temperature would correspond to a resistance of \(51.47 \Omega\) on this thermometer?

The issue of global warming is receiving considerable attention these days. Write a technical report on the subject of global warming. Explain what is meant by the term global warming and discuss objectively the scientific evidence that is cited as the basis for the argument that global warming is occurring.

The following table lists temperatures and specific volumes of water vapor at two pressures: \begin{tabular}{ccccc} \hline\(p=1.0 \mathrm{MPa}\) & & \multicolumn{2}{c}{\(p=1.5 \mathrm{Mpa}\)} \\ \cline { 1 - 2 } \hline\(\left({ }^{\circ} \mathrm{C}\right)\) & \(v\left(\mathrm{~m}^{3} / \mathrm{kg}\right)\) & & \(T\left({ }^{\circ} \mathrm{C}\right)\) & \(v\left(\mathrm{~m}^{3} / \mathrm{kg}\right)\) \\ \hline 200 & \(0.2060\) & 200 & \(0.1325\) \\ 240 & \(0.2275\) & 240 & \(0.1483\) \\ 280 & \(0.2480\) & 280 & \(0.1627\) \\ \hline \end{tabular} Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate (a) the specific volume at \(T=240^{\circ} \mathrm{C}, p=1.25 \mathrm{MPa}\), in \(\mathrm{m}^{3} / \mathrm{kg}\). (b) the temperature at \(p=1.5 \mathrm{MPa}, v=0.1555 \mathrm{~m}^{3} / \mathrm{kg}\), in \({ }^{\circ} \mathrm{C}\). (c) the specific volume at \(T=220^{\circ} \mathrm{C}, p=1.4 \mathrm{MPa}\), in \(\mathrm{m}^{3} / \mathrm{kg}\).
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