/*! 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 4 A man slowly pulls a bucket of w... [FREE SOLUTION] | 91Ó°ÊÓ

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A man slowly pulls a bucket of water from a well of depth \(h=20 \mathrm{~m}\). The mass of the uniform rope and bucket full of water are \(m=200 \mathrm{~g}\) and \(M 19.9 \mathrm{~kg}\), respectively. Find the work done (in kJ) by the man.

Short Answer

Expert verified
The work done is approximately 3.94 kJ.

Step by step solution

01

Understand the Problem

We want to find the work done in pulling a bucket of water from the bottom of a well. The depth of the well, the mass of the bucket with water, and the mass of the rope are given. The work is done against gravity.
02

Convert Units

Convert the mass of the rope from grams to kilograms. Since there are 1000 grams in a kilogram, the mass of the rope is \(m = \frac{200}{1000} = 0.2 \, \mathrm{kg}\).
03

Use Work Formula

Work done against gravity is given by the formula \(W = F \cdot d\), where \(F\) is the force (weight) and \(d\) is the distance (depth of the well). Total force is the weight of the bucket plus the weight of the rope, \(F = (M + m)g\), where \(g = 9.81 \, \mathrm{m/s^2}\). The depth \(d\) is 20 m.
04

Calculate Total Force

The total mass of the bucket and rope is \(M + m = 19.9 + 0.2 = 20.1 \, \mathrm{kg}\). The force due to this mass is \[ F = 20.1 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} = 197.181 \, \mathrm{N}. \]
05

Calculate Work Done

Now, use the work formula with distance of 20 m:\[ W = 197.181 \, \mathrm{N} \times 20 \, \mathrm{m} = 3943.62 \, \mathrm{J}. \]
06

Convert Work to kJ

Convert the work from Joules to kilojoules by dividing by 1000: \[ W = \frac{3943.62}{1000} = 3.94362 \, \mathrm{kJ}. \]

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Key Concepts

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

Understanding Gravity
Gravity is the force that pulls objects towards the center of the Earth. It's the reason why we stay grounded and why objects fall when dropped. The strength of gravity on Earth is constant, approximately \( g = 9.81 \ \mathrm{m/s^2} \). This value represents the acceleration due to gravity. When calculating forces or any physical effort involving vertical movement on Earth, understanding this gravitational pull is essential.
For instance, in this problem, when the man pulls the bucket, he is working against gravity. This means he must exert a force equal to the gravitational pull on the bucket and the rope. The concept is simple: the greater the mass, the greater the force needed due to gravity. To calculate the weight or force needed to lift any object, multiply its mass by \( g \). This calculation helps you understand how much effort or work must be put in fora given task that involves lifting or moving objects vertically.
Calculating Force
Force is a push or pull upon an object resulting from its interaction with another object. In this exercise, we calculate force using the formula \( F = (M + m)g \), where \( M \) is the mass of the bucket with water, \( m \) is the mass of the rope, and \( g \) is gravity.

Steps in Force Calculation

  • First, add the masses of all objects involved (bucket and rope) to find the total mass.
  • Use the formula \( F = (M + m)g \). This gives the force as the weight of the combined mass.
  • For example, in this exercise, you calculate \( 20.1 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} = 197.181 \, \mathrm{N} \) as the total force.
Understanding the force calculation helps in determining how much force must be exerted to perform any work, such as pulling the bucket in our problem. This provides insight into the physical effort required in everyday tasks involving mass and gravity.
Basics of Unit Conversion
Unit conversion is critical when calculating measurements, as it ensures that all values are in the correct units before starting calculations. In this exercise, unit conversion is used to convert the mass of a rope from grams to kilograms because the standard unit for mass in physics equations is kilograms.

How to Convert Units

  • Identify the units you have and the units you need. Here, we're converting grams to kilograms.
  • Use the conversion factor. Since \( 1 \ \mathrm{kg} = 1000 \ \mathrm{g} \), divide the grams by 1000 to get kilograms.
  • In our example, convert \( 200 \, \mathrm{g} \to 0.2 \, \mathrm{kg} \), making it compatible with the rest of the calculations.
Ensuring that calculations involve consistent units prevents errors and provides accurate results. Once all units are properly converted, one can confidently move forward with calculations like those for mechanical work or force.

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Most popular questions from this chapter

One end of an unstretched vertical spring is attached to the ceiling and an object attached to the other end is slowly lowered to its equilibrium position. If \(S\) is the gain in spring energy and \(G\) is the loss in gravitational potential energy in the process, then (1) \(S=G\) (2) \(S=2 G\) (3) \(G=2 S\) (4) None of these

A small block of mass \(2 \mathrm{~kg}\) is kept on a rough inclined surface of inclination \(\theta=30^{\circ}\) fixed in a lift. The lift goes up with a uniform speed of \(1 \mathrm{~ms}^{-1}\) and the block does not slide relative to the inclined surface. The work done by the force of friction on the block in a time interval of \(2 \mathrm{~s}\) is (1) Zero (2) \(9.8 \mathrm{~J}\) (3) 294 I (4) \(16.9 \mathrm{~J}\)

A particle is taken from point \(A\) to point \(B\) under the influence of a force field. Now it is taken back from \(B\) to \(A\) and it is observed that the work done in taking the particle from \(A\) to \(B\) is not equal to the work done in taking it from \(B\) to \(A\). If \(W_{n c}\) and \(W_{c}\) are the work done by non- conservative and conservative forces present in the system, respectively \(\Delta U\) is the change in potential energy and \(\Delta k\) is the chan in kinetic energy, then (1) \(W_{n c}-\Delta U=\Delta k\) (2) \(W_{e}=-\Delta U\) (3) \(W_{n c}^{n}+W_{c}=\Delta k\) (4) \(W_{n c}-\Delta U=-\Delta k\)

Mark the correct statement(s). (1) Total work done by internal forces of a system on the system is always zero. (2) Total work done by internal forces of a system on the system is sometimes zero. (3) Total work done by internal forces acting between the particles of a rigid body is always zero. (4) Total work done by internal forces acting between the particles of a rigid body is soinetimes zero.

A machine delivers power to a body which is proportional to velocity of the body. If the body starts with a velocity which is almost negligible, then the distance covered by the body is proportional to (1) \(\sqrt{v}\) (2) \(\sqrt[3]{\frac{v}{2}}\) (3) \(v^{5 / 3}\) (4) \(v^{2}\)

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