/*! 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 7 Jennifer lifts a 2.5 -kg carton ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Jennifer lifts a 2.5 -kg carton of cat litter from the floor to a height of \(0.75 \mathrm{m} .\) (a) How much total work is done on the carton during this operation? Jennifer then pours \(1.2 \mathrm{kg}\) of the litter into the cat's litter box on the floor. (b) How much work is done by gravity on the $1.2 \mathrm{kg}$ of litter as it falls into the litter box?

Short Answer

Expert verified
Question: Calculate the work done by Jennifer on lifting the carton and the work done by gravity on the litter falling into the box. Answer: (a) Work done on the carton by Jennifer: (2.5 kg x 9.81 m/s^2) x 0.75 m, (b) Work done by gravity on the litter falling into the litter box: (1.2 kg x 9.81 m/s^2) x 0.75 m.

Step by step solution

01

(a) Work done on the carton by Jennifer

To find the work done on the carton by Jennifer while lifting it to a height of 0.75 m, we first need to calculate the force exerted by her during the process, and for that, we will use the gravitational force formula. Force = mass x acceleration due to gravity Force = 2.5 kg x 9.81 m/s^2 Now we have the force exerted. The work done is given by the formula: Work done = force x displacement x cos(theta) Since Jennifer is lifting the carton vertically, the angle (theta) between the force and displacement is 0 degrees. So, cos(theta) = cos(0) = 1. Work done = Force x Displacement x 1 (since cos(0) = 1) Work done = (2.5 kg x 9.81 m/s^2) x 0.75 m
02

(b) Work done by gravity on the litter falling into the litter box

As the litter falls, we need to find the work done by gravity on the 1.2 kg of litter. Since the falling motion is vertical as well, the angle (theta) between the force and displacement is 0 degrees (cos(theta) = 1). We need to find the force exerted by gravity on the 1.2 kg of litter: Force = mass x acceleration due to gravity Force = 1.2 kg x 9.81 m/s^2 For the work done, we need to find the displacement as well. As the litter falls from the 0.75 m height, this is the displacement over which gravity acts: Displacement = 0.75 m Now, we can calculate the work done by gravity: Work done = force x displacement x cos(theta) Work done = (1.2 kg x 9.81 m/s^2) x 0.75 m x 1 (since cos(0) = 1)

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!

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 hang glider moving at speed \(9.5 \mathrm{m} / \mathrm{s}\) dives to an altitude 8.2 m lower. Ignoring drag, how fast is it then moving?

The orbit of Halley's comet around the Sun is a long thin ellipse. At its aphelion (point farthest from the Sun), the comet is $5.3 \times 10^{12} \mathrm{m}\( from the Sun and moves with a speed of \)10.0 \mathrm{km} / \mathrm{s} .$ What is the comet's speed at its perihelion (closest approach to the Sun) where its distance from the Sun is \(8.9 \times 10^{10} \mathrm{m} ?\)
Lars, of mass \(82.4 \mathrm{kg},\) has been working out and can do work for about 2.0 min at the rate of \(1.0 \mathrm{hp}(746 \mathrm{W})\) How long will it take him to climb three flights of stairs, a vertical height of $12.0 \mathrm{m} ?$
A brick of mass \(1.0 \mathrm{kg}\) slides down an icy roof inclined at \(30.0^{\circ}\) with respect to the horizontal. (a) If the brick starts from rest, how fast is it moving when it reaches the edge of the roof $2.00 \mathrm{m}$ away? Ignore friction. (b) Redo part (a) if the coefficient of kinetic friction is $0.10.
A block of wood is compressed 2.0 nm when inward forces of magnitude $120 \mathrm{N}$ are applied to it on two opposite sides. (a) Assuming Hooke's law holds, what is the effective spring constant of the block? (b) Assuming Hooke's law still holds, how much is the same block compressed by inward forces of magnitude \(480 \mathrm{N} ?\) (c) How much work is done by the applied forces during the compression of part (b)?
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.