/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A weight lifter lifts a \(350-\m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A weight lifter lifts a \(350-\mathrm{N}\) set of weights from ground level to a position over his head, a vertical distance of \(2.00 \mathrm{~m}\). How much work does the weight lifter do, assuming he moves the weights at constant speed?

Short Answer

Expert verified
The weightlifter does 700 Joules of work to lift the weights over his head.

Step by step solution

01

Identify given data

Read the problem carefully and identify the known quantities. Here, we know that the Force = 350 N and the Distance = 2.00 m.
02

Write down the formula for work

The formula for computing work is given by \( Work = Force \times Distance \), where force is measured in Newton and distance in metres.
03

Substitute the given values into the formula

Substitute the values of force and distance into the equation. We get Work = 350 N x 2.00 m.
04

Compute the work

By multiplying 350 N and 2.00 m, we find that the work done is 700 J (Joules).

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Force
In physics, force is a push or pull upon an object resulting from the object's interaction with another object. When there's an interaction between two objects, there is a force upon each of the objects. Whenever the interaction ceases, the two objects no longer experience the force. We use Newtons (N) as the unit of force, named after Sir Isaac Newton.
When lifting weights, the weight lifter needs to apply a force to counteract the force of gravity. In our exercise, the force applied is 350 Newtons. These 350 Newtons are used to stabilize the weights over the weight lifter's head at constant speed. This means the force applied balances out the gravitational force acting downwards on the weights, allowing for steady and safe elevation without acceleration.
Some key points to remember about force:
  • It is a vector quantity, which means it has both magnitude and direction.
  • The greater the force applied, the greater the potential for movement or deformation of the object.
  • Force needs to be applied over a distance for work to be done, which brings us to the concept of work in physics.
Covering Distance
Distance, in physics, is the measure of how much ground has been covered during motion. It is a scalar quantity, meaning it only has magnitude and no direction. In our problem, the weight lifter moves the weights a vertical distance of 2.00 meters. This is the displacement the weights experience away from the starting point on the ground level.
When considering work, the distance over which the force is applied is crucial. Without movement across a distance, no work is done, even if a force is applied. For instance:
  • If the weight lifter simply holds the weights steady overhead without moving them, technically no work is done since there's no distance covered in the direction of the force.
  • The direction of the distance moved matters; for work to be calculated, the movement needs to be in the same direction as the applied force.
  • Distance is usually measured in meters (m), consistent with the metric system you'll use for all physics calculations.
Measuring Work Done in Joules
In physics, work is the measure of energy transfer that occurs when an object is moved over a distance by an external force. Work is defined mathematically as the product of force and distance through the equation: \( Work = Force \times Distance \).The result of this calculation gives us work in Joules, with 1 Joule (J) being equivalent to a force of one Newton moving an object over a distance of one meter.
So, in our exercise:
  • The weight lifter applies a force of 350 N over a distance of 2.00 m, resulting in work done of 700 J.
  • The calculation looks like this: \( Work = 350 \text{ N} \times 2.00 \text{ m} = 700 \text{ J} \).
  • Kinetic energy is transferred from the lifter to the weights, raising them to the new position, thus the work done is positive.
Simply put, Joules provide a measure for quantifying how much energy is transferred during the process, helping us understand how much effort is involved in tasks like lifting weights.

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 2.0-m-long pendulum is released from rest when the support string is at an angle of \(25^{\circ}\) with the vertical. What is the speed of the bob at the bottom of the swing?

The masses of the javelin, discus, and shot are \(0.80 \mathrm{~kg}, 2.0 \mathrm{~kg}_{1}\) and \(7.2 \mathrm{~kg}\), respectively, and record throws in the corresponding track events are about \(98 \mathrm{~m}, 74 \mathrm{~m}_{1}\) and \(23 \mathrm{~m}\), respectively. Neglecting air resistance, (a) calculate the minimum initial kinetic energies that would produce these throws, and (b) estimate the average force exerted on each object during the throw, assuming the force acts over a distance of \(2.0 \mathrm{~m} .\) (c) Do your results suggest that air resistance is an important factor?

A truck travels uphill with constant velocity on a highway with a \(7.0^{\circ}\) slope. A 50 -kg package sits on the floor of the back of the truck and does not slide, due to a static frictional force. During an interval in which the truck travels \(340 \mathrm{~m}\), what is the net work done on the package? What is the work done on the package by the force of gravity, the normal force, and the friction force?

A roller-coaster car of mass \(1.50 \times 10^{4} \mathrm{~kg}\) is initially at the top of a rise at point \((1)\). It then moves \(35.0 \mathrm{~m}\) at an angle of \(50.0^{\circ}\) below the horizontal to a lower point (B). (a) Find both the potential energy of the system when the car is at points ( 8 and \((\) B) and the change in potential energy as the car moves from point (A) to point (B), assuming \(y=0\) at point (B). (b) Repeat part (a), this time choosing \(y=0\) at point \(\mathbb{C}\), which is another \(15.0 \mathrm{~m}\) down the same slope from point (B),

A skier of mass \(70 \mathrm{~kg}\) is pulled up a slope by a motordriven cable. (a) How much work is required to pull him \(60 \mathrm{~m}\) up a \(30^{\circ}\) slope (assumed frictionless) at a constant speed of \(2.0 \mathrm{~m} / \mathrm{s} ?\) (b) What power must a motor have to perform this task?

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.