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Chapter 7: Problem 56

An initially stationary \(2.0 \mathrm{~kg}\) object accelerates horizontally and uniformly to a speed of \(10 \mathrm{~m} / \mathrm{s}\) in \(3.0 \mathrm{~s}\). (a) In that \(3.0 \mathrm{~s}\) interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?

Short Answer

Expert verified
(a) 100 J; (b) 66.6 W; (c) 33.3 W.

Step by step solution

01

Determine the acceleration

Since the object starts from rest and reaches a speed of \(10 \mathrm{~m/s}\) in \(3.0 \mathrm{~s}\), we can use the formula for acceleration: \[ a = \frac{v - u}{t} \] where \(v = 10 \, \mathrm{m/s}\) is the final velocity, \(u = 0 \, \mathrm{m/s}\) is the initial velocity, and \(t = 3.0 \, \mathrm{s}\) is the time. Calculating the acceleration, \[ a = \frac{10 \mathrm{~m/s} - 0 \mathrm{~m/s}}{3.0 \mathrm{~s}} = \frac{10}{3} \mathrm{~m/s}^2 \approx 3.33 \mathrm{~m/s}^2 \]
02

Calculate the force on the object

Using Newton鈥檚 second law \( F = ma \), where \( m = 2.0 \mathrm{~kg}\) is the mass and \( a = 3.33 \mathrm{~m/s}^2 \) is the acceleration, we find the force: \[ F = 2.0 \mathrm{~kg} \times 3.33 \mathrm{~m/s}^2 = 6.66 \mathrm{~N} \]
03

Find the distance traveled

Use the formula for distance traveled under uniform acceleration: \[ s = ut + \frac{1}{2} a t^2 \] Since the initial velocity \( u = 0 \mathrm{~m/s} \), the formula simplifies to: \[ s = \frac{1}{2} \times 3.33 \mathrm{~m/s}^2 \times (3.0 \mathrm{~s})^2 \] \[ s = 0.5 \times 3.33 \times 9.0 = 14.985 \mathrm{~m} \approx 15.0 \mathrm{~m} \]
04

Calculate the work done

Work done by the force is given by \( W = F \times s \), where \( F = 6.66 \mathrm{~N} \) and \( s = 15.0 \mathrm{~m} \). So, \[ W = 6.66 \mathrm{~N} \times 15.0 \mathrm{~m} = 99.9 \mathrm{~J} \approx 100 \mathrm{~J} \]
05

Determine the instantaneous power at the end of interval

Power is the rate of doing work, calculated by \( P = Fv \) where \( v = 10 \mathrm{~m/s} \) is the velocity at the end of the interval. Thus, \[ P = 6.66 \mathrm{~N} \times 10 \mathrm{~m/s} = 66.6 \mathrm{~W} \]
06

Compute instantaneous power at halfway

At halfway mark \( t = 1.5 \mathrm{~s} \), velocity \( v \) can be found using \( v = u + at \):\[ v = 0 + 3.33 \frac{\mathrm{m}}{\mathrm{s}^2} \times 1.5 \mathrm{~s} = 5 \mathrm{~m/s} \] The instantaneous power is then \( P = Fv = 6.66 \mathrm{~N} \times 5 \mathrm{~m/s} = 33.3 \mathrm{~W} \).

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a cornerstone of classical mechanics and provides the relationship between force, mass, and acceleration. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. This can be expressed mathematically as \( F = ma \). Here, \( F \) is the force applied, \( m \) is the mass of the object, and \( a \) is the acceleration.

In our example, a 2.0 kg object accelerates to a speed of 10 m/s in 3.0 seconds. We first find the acceleration using the formula \( a = \frac{v - u}{t} \), resulting in an acceleration of approximately 3.33 m/s虏. Applying Newton's Second Law, the force required for this acceleration is calculated as \( 6.66 \) N.

This reveals how an object's motion can be precisely controlled by altering the force applied to it, given its mass and desired acceleration. Understanding this law is crucial as it enables us to predict how objects will respond to various force inputs.
Work-Energy Principle
The work-energy principle connects the concept of work with the change in energy of an object. It states that the work done on an object is equal to the change in its kinetic energy. The formula for work is \( W = F \times s \), where \( F \) is the force applied and \( s \) is the distance over which the force is applied.

In our exercise, the object moves a distance of 15.0 m under the force of 6.66 N. Therefore, the work done on the object is calculated to be approximately 100 J. This work done by the force causes a change in the kinetic energy of the object, allowing it to reach the speed of 10 m/s from rest.

When studying mechanics, always remember that the work-energy principle provides a clear way to understand how forces influence the energy states of physical systems.
Power in Physics
Power in Physics refers to the rate at which work is done or energy is transferred. It is a crucial concept because it not only considers the amount of work but also how fast it is completed. The formula for power is \( P = Fv \), where \( F \) is the force and \( v \) is the velocity of the moving object.

In this scenario, at the end of the 3-second interval, the object's velocity is 10 m/s, and the power generated due to the accelerating force is 66.6 W. This means that 66.6 joules of energy are being transferred every second to the object by the force.

Moreover, it's interesting to see that at the midpoint, when the object has a velocity of 5 m/s, the power is half of the final power, 33.3 W. This exemplifies how varying speeds and durations impact the power requirement.

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Most popular questions from this chapter

The loaded cab of an elevator has a mass of \(3.0 \times 10^{3} \mathrm{~kg}\) and moves \(210 \mathrm{~m}\) up the shaft in \(23 \mathrm{~s}\) at constant speed. At what average rate does the force from the cable do work on the cab?

A coin slides over a frictionless plane and across an \(x y\) coordinate system from the origin to a point with \(x y\) coordinates \((3.0 \mathrm{~m}, 4.0 \mathrm{~m})\) while a constant force acts on it. The force has magnitude \(2.0 \mathrm{~N}\) and is directed at a counterclockwise angle of \(100^{\circ}\) from the positive direction of the \(x\) axis. How much work is done by the force on the coin during the displacement?

A constant force of magnitude \(10 \mathrm{~N}\) makes an angle of \(150^{\circ}\) (measured counterclockwise) with the positive \(x\) direction as it acts on a \(2.0 \mathrm{~kg}\) object moving in an \(x y\) plane. How much work is done on the object by the force as the object moves from the origin to the point having position vector \((2.0 \mathrm{~m}) \hat{\mathrm{i}}-(4.0 \mathrm{~m}) \hat{\mathrm{j}} ?\)

The force on a particle is directed along an \(x\) axis and given by \(F=F_{0}\left(x / x_{0}-1\right)\). Find the work done by the force in moving the particle from \(x=0\) to \(x=2 x_{0}\) by (a) plotting \(F(x)\) and measuring the work from the graph and (b) integrating \(F(x)\).

An explosion at ground level leaves a crater with a diameter that is proportional to the energy of the explosion raised to the \(\frac{1}{3}\) power; an explosion of 1 megaton of TNT leaves a crater with a \(1 \mathrm{~km}\) diameter. Below Lake Huron in Michigan there appears to be an ancient impact crater with a \(50 \mathrm{~km}\) diameter. What was the kinetic energy associated with that impact, in terms of (a) megatons of TNT (1 megaton yields \(\left.4.2 \times 10^{15} \mathrm{~J}\right)\) and (b) Hiroshima bomb equivalents (13 kilotons of TNT each)? (Ancient meteorite or comet impacts may have significantly altered Earth's climate and contributed to the extinction of the dinosaurs and other life-forms.)
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