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Chapter 8: Problem 20

A person is painting his house walls. He stands on a ladder with a bucket containing paint in one hand and a brush in other. Suddenly the bucket slips from his hand and falls down on the floor. If the bucket with the paint had a mass of \(6 \cdot 0 \mathrm{~kg}\) and was at a height of \(2 \cdot 0 \mathrm{~m}\) at the time it slipped, how much gravitational potential energy is lost together with the paint?

Short Answer

Expert verified
The gravitational potential energy lost is 117.6 Joules.

Step by step solution

01

Understand the Formula for Potential Energy

Gravitational potential energy is calculated using the formula \( PE = m \, g \, h \). Here, \( m \) represents the mass of the object, \( g \) is the acceleration due to gravity, and \( h \) is the height of the object above the ground.
02

Identify the Given Values

From the problem, we know the mass \( m = 6.0 \, \text{kg} \), the height \( h = 2.0 \, \text{m} \), and the standard acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \).
03

Substitute the Values into the Formula

Substitute \( m = 6.0 \), \( g = 9.8 \), and \( h = 2.0 \) into the potential energy formula: \[ PE = 6.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2.0 \, \text{m} \].
04

Calculate the Potential Energy

Calculate the potential energy: \( 6.0 \times 9.8 \times 2.0 = 117.6 \, \text{Joules} \). So, the gravitational potential energy is 117.6 Joules.

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Key Concepts

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

Mass
Mass is a fundamental concept in physics that defines the quantity of matter in an object. It is a scalar quantity, which means it only has a magnitude and no direction. The mass of an object has a direct impact on its gravitational attraction to other objects. In most practical cases, mass is measured in kilograms (kg).

In the exercise described, the paint bucket with the paint has a mass of 6.0 kg. This mass determines how much gravitational force the Earth exerts on it. Importantly, more mass means the object will have more potential energy when lifted to a certain height, if other factors remain constant. Mass is crucial in calculating gravitational potential energy because potential energy is directly proportional to the mass of the object.
Acceleration due to gravity
Acceleration due to gravity is a universal constant denoted by the symbol \( g \). On Earth, this value is approximately \( 9.8 \, \text{m/s}^2 \). This constant value represents how quickly an object's velocity increases as it falls due to Earth's gravitational pull.

In the given exercise, the constant \( g = 9.8 \, \text{m/s}^2 \) was used to calculate gravitational potential energy. This value of \( g \) is the average acceleration objects experience when falling freely towards the Earth's surface. It's important because it helps determine how much gravitational pull is affecting the object, which in turn influences the potential energy calculation.

Understanding acceleration due to gravity facilitates comprehension of many physical phenomena, including why objects fall at the rates they do and how energy transformations occur between potential and kinetic energy.
Energy calculation
Energy calculation, in this context, refers to the calculation of gravitational potential energy. This is done using the formula \( PE = m \, g \, h \), where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is the height above the ground.

In the exercise, calculation involves substituting the known values into the formula: a mass \( m = 6.0 \, \text{kg} \), acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \), and height \( h = 2.0 \, \text{m} \). Following these substitutions, calculating \( 6.0 \times 9.8 \times 2.0 \) results in 117.6 Joules.

This calculated energy signifies the amount of potential energy lost as the paint bucket falls. Knowing how to perform this calculation helps in understanding energy transformations, as when the bucket falls, its potential energy converts to kinetic energy until it impacts the ground.

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Most popular questions from this chapter

A block of mass \(5^{\circ} 0 \mathrm{~kg}\) slides down an incline of inclination \(30^{\circ}\) and length \(10 \mathrm{~m}\). Find the work done by the force of gravity.

A particle moves from a point \(\overrightarrow{r_{1}}=(2 \mathrm{~m}) \vec{i}+(3 \mathrm{~m}) \vec{j}\) to another point \(\overrightarrow{r_{2}}=(3 \mathrm{~m}) \vec{i}+(2 \mathrm{~m}) \vec{j}\) during which a certain force \(\vec{F}=(5 \mathrm{~N}) \vec{i}+(5 \mathrm{~N}) \vec{j}\) acts on it. Find the work done by the force on the particle during the displacement.

A small block of mass \(200 \mathrm{~g}\) is kept at the top of a frictionless incline which is \(10 \mathrm{~m}\) long and \(3 \cdot 2 \mathrm{~m}\) high. How much work was required (a) to lift the block from the ground and put it at the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground, if (c) it falls off the incline and drops vertically on the ground (d) it slides down the incline ? Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\)

A man moves on a straight horizontal road with a block of mass \(2 \mathrm{~kg}\) in his hand. If he covers a distance of \(40 \mathrm{~m}\) with an acceleration of \(0.5 \mathrm{~m} / \mathrm{s}^{2}\), find the work done by the man on the block during the motion.

A block of mass \(2 \cdot 0 \mathrm{~kg}\) is pushed down an inclined plane of inclination \(37^{\circ}\) with a force of \(20 \mathrm{~N}\) acting parallel to the incline. It is found that the block moves on the incline with an acceleration of \(10 \mathrm{~m} / \mathrm{s}^{2}\). If the block started from rest, find the work done (a) by the applied force in the first second, (b) by the weight of the block in the first second and (c) by the frictional force acting on the block in the first second. Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\).
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