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Chapter 6: Problem 31

A 4.0-kg object is lifted \(1.5 \mathrm{~m}\). (a) How much work is done against the Earth's gravity? (b) Repeat if the object is lowered instead of lifted.

Short Answer

Expert verified
(a) 58.8 J; (b) -58.8 J.

Step by step solution

01

Understanding Work Against Gravity

When an object is lifted, work is done against Earth's gravity. The amount of work done can be calculated using the formula: \[ W = F imes d \]where \( W \) is the work done, \( F \) is the force, and \( d \) is the distance moved in the direction of the force.
02

Calculate Force Due to Gravity

The force required to lift the object can be determined by calculating the gravitational force on the object. This force is given by:\[ F = m imes g \]where \( m = 4.0 \text{ kg} \) is the mass and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.Substituting the values:\[ F = 4.0 \text{ kg} imes 9.8 \text{ m/s}^2 = 39.2 \text{ N} \]
03

Calculate Work Done When Lifted

Using the force calculated and the height lifted, the work done against gravity can be found using:\[ W = F imes d = 39.2 \text{ N} imes 1.5 \text{ m} \]\[ W = 58.8 \text{ J} \]So, the work done in lifting the object is 58.8 Joules.
04

Understanding Work When Lowering

When the object is lowered, gravity does the work on the object. The magnitude of work done by gravity is the same as when lifting the object, but the direction is opposite. Work done by the object is negative in the case of lowering.
05

Calculate Work Done When Lowered

Using the same formula and values:\[ W = F imes d = 39.2 \text{ N} imes 1.5 \text{ m} \]\[ W = -58.8 \text{ J} \]Thus, the work done by the object when lowering (work done on the object by gravity) is -58.8 Joules.

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Key Concepts

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

Gravitational Force
Gravitational force is the force of attraction between two objects with mass. It is a fundamental force in nature and is what keeps us grounded on the Earth. The formula to calculate the gravitational force acting on an object near the Earth's surface is:
  • \[ F = m \times g \]
where:
  • \( m \) is the mass of the object (measured in kilograms, kg).
  • \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth's surface).
This force is often called the weight of the object. For instance, if you have a 4.0-kg object, its weight, which is the gravitational force acting upon it, would be 39.2 Newtons. This weight is crucial for determining how much effort is needed to lift the object against Earth's gravity.
Calculation of Work
Work, in the context of physics, is the measure of energy transfer that occurs when an object is moved over a distance by an external force. For work to occur, the force and the movement must be in the same direction. Work is calculated using:
  • \[ W = F \times d \]
where:
  • \( W \) is the work done (measured in Joules, J).
  • \( F \) is the force exerted (in Newtons, N).
  • \( d \) is the distance moved in the direction of the force (in meters, m).
When lifting an object, work is done against gravity. For example, to lift a 4.0-kg object 1.5 meters high, the work done is the gravitational force times the height lifted. As calculated, the necessary work is 58.8 Joules. When the object is lowered back, the same amount of work is transferred back into the system, but in the opposite direction, signifying negative work.
Physics Problem Solving
Solving physics problems often involves applying the right formulas and understanding concepts like force and work. To solve such problems, consider these steps:
  • Identify what you know: Read the problem carefully, and determine what information is given. In our case, the object's mass (4.0 kg) and movement (1.5 m up or down) are provided.
  • Determine what you need to find: Look for what the problem asks. Here, the task is to find the work done lifting and lowering an object.
  • Apply relevant physics concepts: Use the formulas for gravitational force \( F = m \times g \) and work \( W = F \times d \). Calculate these separately for lifting and lowering.
  • Solve using calculations: Plugin the numbers from the problem into the equations. Verify if they make sense physically (e.g., work should be positive for lifting and negative for lowering).
This systematic approach aids in understanding and solving problems logically, ensuring comprehensive learning of physics concepts.

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Most popular questions from this chapter

How much work is done against gravity in lifting a \(3.0\) -kg object through a vertical distance of \(40 \mathrm{~cm}\) ? An external force is needed to lift an object. If the object is raised at constant speed, the lifting force must equal the weight of the object. The work done by the lifting force is referred to as work done against gravity. Because the lifting force is \(m g\), where \(m\) is the mass of the object, Work \(=(m g)(h)(\cos \theta)=(3.0 \mathrm{~kg} \times 9.81 \mathrm{~N})(0.40 \mathrm{~m})(1)=12 \mathrm{~J}\) In general, the work done against gravity in lifting an object of mass \(m\) through a vertical distance \(h\) is \(m g h\).

A moving 300 -g object slides unpushed \(80 \mathrm{~cm}\) in a straight line along a horizontal tabletop. How much work is done in overcoming friction between the object and the table if the coefficient of kinetic friction is \(0.20\) ? First find the friction force. Since the normal force equals the weight of the object, $$ F_{\mathrm{f}}=\mu_{k} F_{N}=(0.20)(0.300 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=0.588 \mathrm{~N} $$ The work done overcoming friction is \(F_{\mathrm{f}} S \cos \theta\). Here \(\theta\) is the angle between the force and the displacement. Because the friction force is opposite in direction to the displacement, \(\theta=180^{\circ}\). Therefore, Work \(=F_{\mathrm{f}} S \cos 180^{\circ}=(0.588 \mathrm{~N})(0.80 \mathrm{~m})(-1)=-0.47 \mathrm{~J}\) The work is negative because the friction force is oppositely directed to the displacement; it slows the object and it decreases the object's kinetic energy, or more to the point, it opposes the motion.

A proton \(\left(m=1.67 \times 10^{-27} \mathrm{~kg}\right)\) that has a speed of \(5.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\) passes through a metal film of thickness \(0.010 \mathrm{~mm}\) and emerges with a speed of \(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). How large an average force opposed its motion through the film?

A 10.0-kg block is launched up a \(30.0^{\circ}\) inclined plane at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). As it slides it loses \(200 \mathrm{~J}\) to friction. How far along the incline will it travel before coming to rest?

A narrowing ladder \(3.0 \mathrm{~m}\) long weighing \(200 \mathrm{~N}\) has its center of gravity \(120 \mathrm{~cm}\) from the bottom. At its top end is a \(50-\mathrm{N}\) can of paint. Compute the work required to raise the ladder from being horizontal, lying on the ground, to being vertical with its legs resting on the ground. In other words, how much work must be done to rotate the ladder into an upright vertical position, thereby raising both its center of gravity and the can of paint? The work done (against gravity) consists of two parts: the work to raise the center of gravity \(1.20 \mathrm{~m}\) and the work to raise the load at the end through \(3.0 \mathrm{~m}\). Therefore, Work done \(=(200 \mathrm{~N})(1.20 \mathrm{~m})+(50 \mathrm{~N})(3.0 \mathrm{~m})=0.39 \mathrm{~kJ}\)
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