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Chapter 3: Problem 42

A certain particle has a weight of \(22 \mathrm{~N}\) at a point where \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\). What are its (a) weight and (b) mass at a point where \(g=4.9 \mathrm{~m} / \mathrm{s}^{2} ?\) What are its (c) weight and (d) mass if it is moved to a point in space where \(g=0 ?\)

Short Answer

Expert verified
a) 10.976 N, b) 2.24 kg, c) 0 N, d) 2.24 kg

Step by step solution

01

- Understand Weight and Mass Relationship

Weight (\text{W}) of an object is given by the product of its mass (\text{m}) and gravitational acceleration (\text{g}). The formula is given by: \[ \text{W} = \text{m} \times \text{g} \]
02

- Calculate Mass at Initial Condition

The initial weight is \( 22 \text{ N} \). Given that \( \text{g} = 9.8 \text{ m/s}^2 \), we can find the mass (\text{m}) using the formula \( \text{W} = \text{m} \times \text{g} \). So, \[ 22 \text{ N} = \text{m} \times 9.8 \text{ m/s}^2 \] Solving for \text{m}: \[ \text{m} = \frac{22}{9.8} \text{ kg} = 2.24 \text{ kg} \].
03

- Calculate Weight at New Gravity

Now that we have the mass, we can find the weight at a point where \( g = 4.9 \text{ m/s}^2 \). Using \( \text{W} = \text{m} \times \text{g} \), \[ \text{W} = 2.24 \text{ kg} \times 4.9 \text{ m/s}^2 = 10.976 \text{ N} \].
04

- Mass at New Gravity

Mass does not depend on gravitational acceleration. Therefore, the mass remains the same, \( \text{m} = 2.24 \text{ kg} \).
05

- Weight in Space

In space where \( g = 0 \): \[ \text{W} = \text{m} \times \text{g} = 2.24 \text{ kg} \times 0 = 0 \text{ N} \].
06

- Mass in Space

Mass remains constant irrespective of the gravitational acceleration. Therefore, the mass is still \( \text{m} = 2.24 \text{ kg} \).

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration is the acceleration of an object caused by the force of gravity from a massive body, typically a planet or star. It is denoted by the symbol \( g \) and measured in meters per second squared (m/s虏).

Here are some critical points to understand:
  • On Earth's surface, the average value of \( g \) is approximately 9.8 m/s虏, but it can vary slightly depending on location.
  • Gravitational acceleration decreases with altitude and increases as you move closer to the planet's core.
  • Different celestial bodies have different values of \( g \), depending on their mass and radius.
When you move to different points with varying gravitational acceleration, the weight of an object changes because weight is a product of mass and gravitational acceleration.

Mass Calculation
Mass is one of the fundamental properties of matter and measures the amount of matter in an object. It remains constant regardless of location or the gravity acting upon it.

The formula to calculate weight is:
\[ \text{W} = \text{m} \times \text{g} \]
Where:
  • \( \text{W} \) is the weight of the object.
  • \( \text{m} \) is the mass.
  • \( \text{g} \) is the gravitational acceleration.
Given the weight (\( 22 \) N) and gravitational acceleration (\( 9.8 \) m/s虏), we solve for mass:
\[ \text{m} = \frac{\text{W}}{\text{g}} = \frac{22 \text{ N}}{9.8 \text{ m/s}^2} = 2.24 \text{ kg} \]
The object鈥檚 mass remains 2.24 kg regardless of any change in gravitational acceleration.

Weight Variation
Weight is a force measured in newtons (N) and is derived from the mass of the object and the gravitational pull it experiences. Weight can vary depending on the strength of the gravitational field.

To illustrate weight variation, let's use different gravitational accelerations:
  • At \( g = 9.8 \text{ m/s}^2 \): Weight is 22 N.
  • At \( g = 4.9 \text{ m/s}^2 \): \( \text{W} = \text{m} \times \text{g} = 2.24 \text{ kg} \times 4.9 \text{ m/s}^2 = 10.976 \text{ N} \).
  • At \( g = 0 \text{ m/s}^2 \): In space, weight becomes zero \( \text{W} = \text{m} \times \text{g} = 2.24 \text{kg} \times 0 = 0 \text{N} \).
It's crucial to note that regardless of the gravitational acceleration, the mass of the object remains constant at 2.24 kg. This separation of mass and weight is fundamental in physics and helps explain how objects behave under different gravitational forces.

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Most popular questions from this chapter

A Navy jet (Fig. 3-34) with a mass of \(2.3 \times 10^{4} \mathrm{~kg}\) requires an airspeed of \(85 \mathrm{~m} / \mathrm{s}\) for liftoff. The engine develops a maximum force of \(1.07 \times\) \(10^{5} \mathrm{~N}\), but that is insufficient for reaching takeoff speed in the \(90 \mathrm{~m}\) runway available on an aircraft carrier. What minimum force (assumed constant) is needed from the catapult that is used to help launch the jet? Assume that the catapult and the jet's engine each exert a constant force over the \(90 \mathrm{~m}\) distance used for takeoff.

A stone is thrown vertically upward. On its way up it passes point \(A\) with speed \(v\), and point \(B, 3.00 \mathrm{~m}\) higher than \(A\), with speed \(\frac{1}{2} v .\) Calculate (a) the speed \(v\) and (b) the maximum height reached by the stone above point \(B\).

\(A\) "sun yacht" is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstances, it can be large enough to send the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of \(900 \mathrm{~kg}\) and reccives a push of \(20 \mathrm{~N}\). (a) What is the magnitude of the resulting acceleration? If the craft starts from rest, (b) how far will it travel in 1 day and (c) how fast will it then be moving?

Water drips from the nozzle of a shower onto the floor \(200 \mathrm{~cm}\) below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. Find the locations of the second and third drops when the first strikes the floor.

Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward force (thrust) of \(3260 \mathrm{~N}\), the craft descends at constant speed; if the engine provides only \(2200 \mathrm{~N}\), the craft accelerates downward at \(0.39\) \(\mathrm{m} / \mathrm{s}^{2}\). (a) What is the weight of the landing craft in the vicinity of Callisto's surface? (b) What is the mass of the craft? (c) What is the magnitude of the free-fall acceleration near the surface of Callisto?
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