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Chapter 5: Problem 79

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
Weight becomes 11 N at \(g = 4.9\) and 0 N at \(g = 0\); mass stays \(2.24 \text{ kg}\) everywhere.

Step by step solution

01

Determine initial mass

Use the weight formula: \( W = m \, g \) to find the mass. Rearranging it gives \( m = \frac{W}{g} \). Initially, \( W = 22 \, \mathrm{N} \) and \( g = 9.8 \, \mathrm{m/s^2} \). So, \( m = \frac{22}{9.8} \approx 2.24 \, \mathrm{kg} \).
02

Calculate weight at new location where \(g = 4.9\, \mathrm{m/s^2}\)

Using the mass found in Step 1, the weight at the new location can be calculated using \( W' = m \times g' \). Substituting, we have \( W' = 2.24 \, \mathrm{kg} \times 4.9 \, \mathrm{m/s^2} \approx 11 \, \mathrm{N} \).
03

Determine mass at new location \(g = 4.9\, \mathrm{m/s^2}\)

Mass does not change with location, so it remains \( 2.24 \, \mathrm{kg} \).
04

Determine weight at location where \(g = 0\)

At \( g = 0 \), the weight is \( W'' = m \times 0 = 0 \, \mathrm{N} \).
05

Determine mass at location where \(g = 0\)

Mass again does not change with location. It remains \( 2.24 \, \mathrm{kg} \).

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Key Concepts

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

Understanding Force
Force is a fundamental concept in physics which describes the interaction that changes the motion of an object. It can cause an object to start moving, stop, or change direction.
A force is usually described by its magnitude (how strong it is) and its direction. The common unit for force is the Newton (N).
  • Forces can be caused by interactions like gravity, friction, or applied force.
  • Force can be calculated using Newton's second law, which states: \( F = m imes a \) where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
In our exercise, the force we're interested in is the gravitational force (weight). This is what pulls the particle downwards due to Earth's gravity.
The Role of Gravity
Gravity is a natural phenomena where all things with mass or energy attract one another. On Earth, gravity gives weight to physical objects and determines the movement of planets and galaxies.
For more practical purposes, gravity causes objects to have weight and accelerates them towards the earth at a rate denoted by \( g \).
  • The standard acceleration due to gravity on Earth is approximately \( 9.8 \, \mathrm{m/s^2} \).
  • Gravity is not constant everywhere. It varies slightly across the Earth and significantly on other celestial bodies or in space.
Thus, when we talk about weight, it directly depends on the local gravitational pull.
Weight Calculation
Weight is the force exerted by gravity on an object and is calculated as the product of its mass and the gravitational acceleration acting on it.
It's expressed using the formula: \( W = m imes g \) where \( W \) is the weight, \( m \) is the mass, and \( g \) is the local gravitational field strength.
  • If an object is taken to a place with different gravitational strength, its weight changes.
  • In our problem, when \( g \) was \( 9.8 \, \mathrm{m/s^2} \), the weight was \( 22 \, \mathrm{N} \). When \( g \) changed to \( 4.9 \, \mathrm{m/s^2} \), the weight found was \( 11 \, \mathrm{N} \).
At a place with no gravitational field (like space where \( g = 0 \)), the weight of any object would be zero.
Mass Calculation
Mass is a measure of the amount of matter in an object. It's an intrinsic property and remains constant regardless of where the object is located.
The relation between mass and weight is given by \( m = \frac{W}{g} \). This formula helps us find the mass of an object when its weight and local gravity are known.
  • In our exercise, the mass was calculated as \( 2.24 \mathrm{kg} \) using the initial conditions.
  • Since mass doesn't depend on gravity, it didn't change even when \( g \) was altered to \( 4.9 \, \mathrm{m/s^2} \) or when \( g \) was zero.
Understanding mass helps us measure how much matter makes up an object and is crucial in calculating how forces like gravity will act on it.

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Most popular questions from this chapter

Tarzan, who weighs \(820 \mathrm{~N}\), swings from a cliff at the end of a \(20.0 \mathrm{~m}\) vine that hangs from a high tree limb and initially makes an angle of \(22.0^{\circ}\) with the vertical. Assume that an \(x\) axis extends horizontally away from the cliff edge and a \(y\) axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is \(760 \mathrm{~N}\). Just then, what are (a) the force on him from the vine in unit- vector notation and the net force on him (b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the \(x\) axis? What are the (e) magnitude and (f) angle of Tarzan's acceleration just then?

Holding on to a towrope moving parallel to a frictionless ski slope, a \(50 \mathrm{~kg}\) skier is pulled up the slope, which is at an angle of \(8.0^{\circ}\) with the horizontal. What is the magnitude \(F_{\text {rope }}\) of the force on the skier from the rope when (a) the magnitude \(v\) of the skier's velocity is constant at \(2.0 \mathrm{~m} / \mathrm{s}\) and (b) \(v=2.0 \mathrm{~m} / \mathrm{s}\) as \(v\) increases at a rate of \(0.10 \mathrm{~m} / \mathrm{s}^{2} ?\)

An elevator cab is pulled upward by a cable. The cab and its single occupant have a combined mass of \(2000 \mathrm{~kg}\). When that occupant drops a coin, its acceleration relative to the cab is \(8.00 \mathrm{~m} / \mathrm{s}^{2}\) downward. What is the tension in the cable?

Using a rope that will snap if the tension in it exceeds \(387 \mathrm{~N}\), you need to lower a bundle of old roofing material weighing \(449 \mathrm{~N}\) from a point \(6.1 \mathrm{~m}\) above the ground. Obviously if you hang the bundle on the rope, it will snap. So, you allow the bundle to accelerate downward. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground?

Compute the weight of a \(75 \mathrm{~kg}\) space ranger (a) on Earth, (b) on Mars, where \(g=3.7 \mathrm{~m} / \mathrm{s}^{2},\) and \((\mathrm{c})\) in interplanetary space, where \(g=0 .\) (d) What is the ranger's mass at each location?
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