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

A dog lifts a \(0.75-\mathrm{kg}\) bone straight up through a distance of \(0.11 \mathrm{~m}\). How much work was done by the dog?

Short Answer

Expert verified
The work done by the dog is 0.8 J.

Step by step solution

01

Identify Given Data

We know the mass of the bone is given as \(m = 0.75\, \text{kg}\) and the distance moved is \(d = 0.11\, \text{m}\). We need to calculate how much work was done by the dog lifting the bone.
02

Understand Work Formula

The work done \(W\) can be calculated using the formula: \(W = F \cdot d\), where \(F\) is the force applied and \(d\) is the distance. The force here is the weight of the bone, which can be calculated using \(F = m \cdot g\), where \(g = 9.8\, \text{m/s}^2\), the acceleration due to gravity.
03

Calculate the Force

The force exerted by the dog is the weight of the bone. Using the formula \(F = m \cdot g\), substitute \(m = 0.75\, \text{kg}\) and \(g = 9.8\, \text{m/s}^2\) to get \(F = 0.75 \times 9.8 = 7.35\, \text{N}\).
04

Calculate the Work Done

Now substitute the values of \(F\) and \(d\) into the work formula. \(W = F \cdot d = 7.35 \times 0.11 = 0.8085\, \text{J} \).
05

Round Off to Appropriate Significant Figures

Considering significant figures from the given data, round the work done to one decimal place: \(W = 0.8\, \text{J}\).

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.

Force
Force is a fundamental concept in physics that describes the interaction which changes an object's state of motion or shape. We often see force in action through pushes, pulls, or lifts. In our bone-lifting scenario, the force exerted by the dog is the bone's weight. The weight is the gravitational force acting on the bone due to Earth's gravity. To calculate this force, you need to know two essential factors:
  • The mass of the object (here, the bone).
  • The acceleration due to gravity, which is the rate of increase in velocity per unit time for a freely falling object.
In our case, the force is calculated as the product of mass and gravitational acceleration: \[ F = m \cdot g \]Given the mass of 0.75 kg and acceleration due to gravity as 9.8 m/s², the force amounts to 7.35 N. This indicates how much force the dog needs to apply to overcome gravity and lift the bone upward.
Acceleration Due to Gravity
The acceleration due to gravity, often represented by the symbol \(g\), is the rate at which objects accelerate as they fall towards Earth. It is approximately 9.8 m/s², varying slightly due to Earth's rotation and shape at different locations. This value is a key component in calculating forces in physics.When considering the lifting of the bone by the dog, \(g\) is used to determine the bone's weight through the formula:\[ F = m \cdot g \]Where:
  • \(m\) is the mass of the object in kilograms.
  • \(g\) is the acceleration due to gravity.
Multiplying the mass (0.75 kg) with \(g\) (9.8 m/s²) provides the gravitational force (or weight) which the dog must counteract to lift the bone.
Significant Figures
Significant figures are important in scientific calculations as they indicate the precision of a measured value. When performing calculations, it's crucial to consider the precision of the given data to ensure results are reported accurately.In our problem, while calculating work, we start with data having two and three significant figures (0.75 kg and 0.11 m, respectively). The calculated force (7.35 N) ends up with three significant figures. When multiplying this force by distance to find work:\[ W = F \cdot d = 7.35 \times 0.11 = 0.8085 \text{ J} \]We follow the rule of using the least number of significant figures (from the given data) in the final answer:
  • 0.75 kg has two significant figures.
  • 0.11 m has two significant figures.
Thus, the work done, rounded to one decimal place to match the initial data's precision, is reported as 0.8 J. This ensures that our answer does not suggest a false degree of accuracy.

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

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