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Chapter 8: Problem 3

You drop a \(2.00 \mathrm{~kg}\) book to a friend who stands on the ground at distance \(D=10.0 \mathrm{~m}\) below. If your friend's outstretched hands are at distance \(d=1.50 \mathrm{~m}\) above the ground (Fig. 8-30), (a) how much work \(W_{g}\) does the gravitational force do on the book as it drops to her hands? (b) What is the change \(\Delta U\) in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy \(U\) of that system is taken to be zero at ground level, what is \(U\) (c) when the book is released and (d) when it reaches her hands? Now take \(U\) to be \(100 \mathrm{~J}\) at ground level and again find (e) \(W_{g},(\mathrm{f}) \Delta U,(\mathrm{~g}) U\) at the release point, and (h) \(U\) at her hands.

Short Answer

Expert verified
(a) 166.6 J. (b) -166.6 J. (c) 196.0 J. (d) 29.4 J. (e) 166.6 J. (f) -166.6 J. (g) 296.0 J. (h) 129.4 J.

Step by step solution

01

Find the total distance fallen

First, we find the total distance the book falls. The book is released from a height of 10.0 m above the ground and stops at 1.5 m above the ground. Therefore, the distance fallen is:\[ d_{\text{fallen}} = D - d = 10.0 \ \text{m} - 1.5 \ \text{m} = 8.5 \ \text{m}. \]
02

Calculate the work done by gravity

The work done by gravity \(W_{g}\) is the product of the gravitational force and the distance fallen in the direction of the force, which is in this case downward:\[ W_{g} = mgh = (2.00 \ \text{kg}) \times (9.8 \ \text{m/s}^2) \times (8.5 \ \text{m}) = 166.6 \ \text{J}, \]where \(m = 2.00 \ \text{kg}\) is the mass, and \(g = 9.8 \ \text{m/s}^2\) is the acceleration due to gravity.
03

Find the change in gravitational potential energy

The change in gravitational potential energy \(\Delta U\) is given by the negative of the work done by gravity, as potential energy decreases when work is done on the object:\[ \Delta U = -W_{g} = -166.6 \ \text{J}. \]
04

Determine the initial gravitational potential energy when released

The gravitational potential energy \(U\) at the initial release point is relative to the ground level, where we assume the potential energy is zero. At the height of 10.0 m, the potential energy is calculated as:\[ U = mgh = (2.00 \ \text{kg}) \times (9.8 \ \text{m/s}^2) \times (10.0 \ \text{m}) = 196.0 \ \text{J}. \]
05

Determine gravitational potential energy at her hands

The gravitational potential energy \(U\) at her hands, 1.5 m above ground, is given by:\[ U = mgh = (2.00 \ \text{kg}) \times (9.8 \ \text{m/s}^2) \times (1.5 \ \text{m}) = 29.4 \ \text{J}. \]
06

Recalculate work done with new reference of zero potential at 100 J

When the potential energy is taken to be 100 J at ground level, the work done by gravity is unchanged, so:\[ W_{g} = 166.6 \ \text{J}. \]
07

Recalculate the change in potential energy with new reference

The change in potential energy \(\Delta U\) remains the same irrespective of the reference point:\[ \Delta U = -166.6 \ \text{J}. \]
08

Recalculate initial potential energy with new reference

With the ground level reference of 100 J, initial potential energy at 10.0 m above is:\[ U = 196.0 \ \text{J} + 100 \ \text{J} = 296.0 \ \text{J}. \]
09

Recalculate potential energy at her hands with new reference

With the reference point having 100 J at ground level, potential energy at her hands 1.5 m above ground is:\[ U = 29.4 \ \text{J} + 100 \ \text{J} = 129.4 \ \text{J}. \]

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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 one of the four fundamental forces of nature. It is a force of attraction that exists between any two masses, no matter how far apart they are. In our scenario, the gravitational force acts between the Earth and the book as you release it. This force causes the book to accelerate towards the ground, paving the way for the work done during its fall.

The gravitational force can be calculated using the formula:
  • Force (\( F_g \)) = mass (\( m \)) 脳 acceleration due to gravity (\( g \)) = (\( 2.00 \, \text{kg} \)) 脳 (\( 9.8 \, \text{m/s}^2 \)).
This formula tells us that the gravitational force acting on the book is a constant force of approximately 19.6 Newtons (N), directed downwards.

Understanding gravitational force is crucial as it is the primary reason why objects fall towards Earth. This force is what does the work when an object is in free fall, contributing to the concepts of work and energy.
Potential Energy
Potential energy is the energy that has the potential to do work due to an object's position or condition. In the case of gravitational potential energy, its potential comes from the height of an object above the ground.

When you hold the book above the ground, it possesses gravitational potential energy due to its height. The gravitational potential energy (\( U \)) at any point can be calculated using the formula:
  • \( U = mgh \)
Where \( m \) is the mass of the book, \( g \) is the acceleration due to gravity, and \( h \) is the height above the reference point.

Initially, when the book is at 10 meters, its potential energy is 196 Joules (\( J \)). When it reaches your friend's hands at 1.5 meters, the potential energy decreases to 29.4 Joules. This decrease represents a change in potential energy as the book converts its potential energy into kinetic energy while falling.
Acceleration Due to Gravity
Acceleration due to gravity (\( g \)) is a measure of how fast an object will accelerate when dropped in a vacuum (where there is no air resistance). On Earth, this value is approximately 9.8 meters per second squared (\( ext{m/s}^2 \)).

This value is constant at different heights because of the massive size of Earth compared to daily objects. It ensures that, regardless of mass, every object will accelerate at the same rate when in free fall. This is crucial for calculations involving gravitational force and potential energy.

Understanding acceleration due to gravity helps in predicting how fast an object will move as it falls. In our exercise, knowing the value of \( g \) allows you to calculate both the work done by gravitational force and the change in potential energy accurately. It provides a reliable base for determining how these energies shift and transform as an object moves relative to gravitational pull.

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Most popular questions from this chapter

At a certain factory, \(300 \mathrm{~kg}\) crates are dropped vertically from a packing machine onto a conveyor belt moving at \(1.20 \mathrm{~m} / \mathrm{s}\) (Fig. 8 -64). (A motor maintains the belt's constant speed.) The coefficient of kinetic friction between the belt and each crate is \(0.400 .\) After a short time, slipping between the belt and the crate ceases, and the crate then moves along with the belt. For the period of time during which the crate is being brought to rest relative to the belt, calculate, for a coordinate system at rest in the factory, (a) the kinetic energy supplied to the crate, (b) the magnitude of the kinetic frictional force acting on the crate, and (c) the energy supplied by the motor. (d) Explain why answers (a) and (c) differ.

A collie drags its bed box across a floor by applying a horizontal force of \(8.0 \mathrm{~N}\). The kinetic frictional force acting on the box has magnitude \(5.0 \mathrm{~N}\). As the box is dragged through \(0.70 \mathrm{~m}\) along the way, what are (a) the work done by the collie's applied force and (b) the increase in thermal energy of the bed and floor?

A \(25 \mathrm{~kg}\) bear slides, from rest, \(12 \mathrm{~m}\) down a lodgepole pine tree, moving with a speed of \(5.6 \mathrm{~m} / \mathrm{s}\) just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

Figure 8 -31 shows a ball with mass \(m=0.341 \mathrm{~kg}\) attached to the end of a thin rod with length \(L=0.452 \mathrm{~m}\) and negligible mass. The other end of the rod is pivoted so that the ball can move in a vertical circle. The rod is held horizontally as shown and then given enough of a downward push to cause the ball to swing down and around and just reach the vertically up position, with zero speed there. How much work is done on the ball by the gravitational force from the initial point to (a) the lowest point, (b) the highest point, and (c) the point on the right level with the initial point? If the gravitational potential energy of the ball-Earth system is taken to be zero at the initial point, what is it when the ball reaches (d) the lowest point, (e) the highest point, and (f) the point on the right level with the initial point? (g) Suppose the rod were pushed harder so that the ball passed through the highest point with a nonzero speed. Would \(\Delta U_{g}\) from the lowest point to the highest point then be greater than, less than, or the same as it was when the ball stopped at the highest point?

A \(4.0 \mathrm{~kg}\) bundle starts up a \(30^{\circ}\) incline with \(128 \mathrm{~J}\) of kinetic energy. How far will it slide up the incline if the coefficient of kinetic friction between bundle and incline is \(0.30 ?\)
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