Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What are the weight in newtons and the mass in kilograms of (a) a 5.00-lb bag of sugar, (b) a 240-lb fullback, and \((c)\) a 1.80-ton car? (1 ton \(=2000\) lb. \()\)

Short Answer

Expert verified
Weight and Mass are: (a) Bag of Sugar: 22.2 N and 2.27 kg, (b) Fullback: 1068 N and 109 kg, (c) Car: 15972 N and 1630 kg

Step by step solution

01

Convert Pounds to Newtons for Each Item

Use the conversion factor of 4.45 newtons equals 1 pound to convert from pounds to newtons. (a) For the 5.00-lb bag of sugar, the weight will be \(5.00 \times 4.45\) N. (b) For the 240-lb fullback, the weight will be \(240 \times 4.45\) N. (c) For the 1.80-ton car, convert tons to pounds, then to newtons. As 1 ton is 2000 lb, the car weighs \( 1.80 \times 2000 \times 4.45\) N.
02

Convert Newtons to Kilograms for Each Item

Substitute the weight in newtons in the following formula to find the mass: \( \text{mass(kg)} = \text{weight(N)} / 9.8 \). (a) For the bag of sugar, the mass will now be \(5.00 \times 4.45 / 9.8\) kg. (b) For the fullback, the mass will be \(240 \times 4.45 / 9.8\) kg. (c) For the car, the mass will be \( 1.80 \times 2000 \times 4.45 / 9.8\) kg.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Newton (N) to Kilograms (kg) conversion
Newton is a unit of force, while kilograms refer to mass. When you have the weight of an object in Newtons, you can convert it to mass in kilograms using Earth's gravity as a conversion factor. On Earth, the standard gravitational acceleration is approximately 9.8 m/s².
To convert a weight in Newtons (N) to mass in kilograms (kg), you use the formula:
  • \[ \text{mass (kg)} = \frac{\text{weight (N)}}{9.8} \]
Suppose you have an object's weight in Newtons. For example, if something weighs 44.5 N on Earth, its mass would be:
  • \[ \text{mass (kg)} = \frac{44.5}{9.8} \approx 4.54 \text{ kg}\]
This conversion is essential for accurately distinguishing between weight (a force) and mass (a property of matter).
Pounds to Newtons conversion
Pounds (lb) are a unit of force commonly used in the United States. When it comes to scientific purposes or international standards, converting to Newtons is necessary. The conversion factor to switch from pounds (lb) to Newtons (N) is approximately 4.45. This means for every pound, you multiply by 4.45 to get the equivalent weight in Newtons.
For example, to convert 5 pounds to Newtons:
  • \[5 \times 4.45 = 22.25 \text{ N} \]
This conversion allows you to communicate weight more universally, given that Newtons are part of the metric system which is widely recognized in scientific communities.
Mass and Weight calculation
Mass and weight are related but distinctly different concepts. Mass measures the amount of matter in an object, and it doesn’t change regardless of the object’s location. Weight, on the other hand, is the force exerted by gravity on an object.
The relationship between weight (in Newtons) and mass (in kilograms) on Earth is given by:
  • Weight(N) = Mass(kg) × Gravity (9.8 m/s²)
Understanding this relationship is crucial, especially when performing calculations and conversions between different units. If you know the weight of an object and Earth’s gravity, you can rearrange this formula to find the mass in kilograms:
  • \[ \text{Mass (kg)} = \frac{\text{Weight (N)}}{9.8} \]
This is helpful in many scientific contexts where knowing the exact mass of an object is necessary for further calculations or experiments.
Metric conversion factors
The metric system is a universal measurement system used widely in science and most countries around the world. Conversion within this system can often involve factors of 10, making it simpler than other systems.
Common metric conversions involve factors like:
  • 1 kilometer = 1000 meters
  • 1 meter = 100 centimeters
  • 1 kilogram = 1000 grams
However, when dealing with force and weight, such as converting pounds to Newtons, it's important to use specific conversion factors like 4.45 to switch from imperial units to metric units effectively.
Utilizing these factors correctly ensures accurate measurements and calculations, making communication of scientific data efficient and less prone to errors.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A neutron travels at a speed of \(1.4 \times 10^{7} \mathrm{~m} / \mathrm{s} .\) Nuclear forces are of very short range, being essentially zero outside a nucleus but very strong inside. If the neutron is captured and brought to rest by a nucleus whose diameter is \(1.0 \times\) \(10^{-14} \mathrm{~m}\), what is the minimum magnitude of the force, presumed to be constant, that acts on this neutron? The neutron's mass is \(1.67 \times 10^{-27} \mathrm{~kg}\)

A car traveling at \(53 \mathrm{~km} / \mathrm{h}\) hits a bridge abutment. A passenger in the car moves forward a distance of \(65 \mathrm{~cm}\) (with respect to the road) while being brought to rest by an inflated air bag. What force (assumed constant) acts on the passenger's upper torso, which has a mass of \(39 \mathrm{~kg}\) ?

How could a 100-lb object be lowered from a roof using a cord with a breaking strength of \(87 \mathrm{lb}\) without breaking the cord?

The Sun yacht Diana, designed to navigate in the solar system using the pressure of sunlight, has a sail area of \(3.1 \mathrm{~km}^{2}\) and a mass of \(930 \mathrm{~kg}\). Near Earth's orbit, the Sun could exert a radiation force of \(29 \mathrm{~N}\) on its sail. (a) What acceleration would such a force impart to the craft? (b) A small acceleration can produce large effects if it acts steadily for a long enough time. Starting from rest then, how far would the craft have moved after 1 day under these conditions? ( \(c\) ) What would then be its speed? (See "The Wind from the Sun," a fascinating science fiction account by Arthur C. Clarke of a Sun yacht race.)

A car moving initially at a speed of \(50 \mathrm{mi} / \mathrm{h}(\approx 80 \mathrm{~km} / \mathrm{h})\) and weighing \(3000 \mathrm{lb}(\approx 13,000 \mathrm{~N})\) is brought to a stop in a distance of \(200 \mathrm{ft}(\approx 61 \mathrm{~m})\). Find \((a)\) the braking force and \((b)\) the time required to stop. Assuming the same braking force, find \((c)\) the distance and \((d)\) the time required to stop if the car were going \(25 \mathrm{mi} / \mathrm{h}(\approx 40 \mathrm{~km} / \mathrm{h})\) initially.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free