/*! 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 9 A box of mass \(6.0 \mathrm{~kg}... [FREE SOLUTION] | 91Ó°ÊÓ

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A box of mass \(6.0 \mathrm{~kg}\) is accelerated from rest by a force across a floor at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) for \(7.0 \mathrm{~s}\). Find the net work done on the box.

Short Answer

Expert verified
The net work done on the box is 588 J.

Step by step solution

01

Understanding the Concepts

To find the net work done on the box, we need to identify the relationship between work, force, and displacement. The net work done on an object can be calculated using the formula for work, which is the product of force and displacement in the direction of the force: \[ W = F \times d \]The force can be calculated using Newton's second law: \[ F = m \times a \]where \( m \) is the mass and \( a \) is the acceleration.
02

Calculate the Force

Use Newton's second law to calculate the force:\[ F = m \times a = 6.0 \text{ kg} \times 2.0 \text{ m/s}^2 \]Calculate the force:\[ F = 12.0 \text{ N} \]
03

Calculate the Displacement

The displacement of the box can be calculated using the equation for displacement with constant acceleration:\[ d = v_0 t + \frac{1}{2} a t^2 \]where \( v_0 = 0 \text{ m/s} \) (since it starts from rest), \( a = 2.0 \text{ m/s}^2 \), and \( t = 7.0 \text{ s} \).Calculate the displacement:\[ d = 0 \times 7 + \frac{1}{2} \times 2.0 \times (7.0)^2 \]\[ d = \frac{1}{2} \times 2.0 \times 49 \]\[ d = 49.0 \text{ m} \]
04

Calculate the Work Done

Now calculate the net work done using the force and displacement:\[ W = F \times d = 12.0 \text{ N} \times 49.0 \text{ m} \]Calculate the work done:\[ W = 588.0 \text{ J} \]

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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 of motion is a fundamental concept in physics that gives us a direct relationship between the force applied to an object, its mass, and the resulting acceleration. While this may sound complex at first, there's a simple equation that ties these ideas together:\[ F = m \times a \]In this equation, \( F \) represents the force in newtons, \( m \) is the mass of the object in kilograms, and \( a \) is the acceleration in meters per second squared. This principle tells us that the force on an object is equal to its mass multiplied by the acceleration that the object is experiencing.
For example, in our exercise, we have a box with a mass of 6.0 kg and it is being accelerated at a rate of 2.0 m/s². By applying Newton's second law, we find that the force is 12.0 N, as calculated by multiplying the mass and acceleration.Newton's second law is instrumental in analyzing and predicting how objects will move under various forces, and acts as a cornerstone of classical mechanics.
Displacement Calculation
When an object moves, it covers a certain distance that can be defined as its displacement. In physics, especially when dealing with constant acceleration, there's a specific formula we use to calculate this displacement:\[ d = v_0 t + \frac{1}{2} a t^2 \]Here, \( d \) is the displacement, \( v_0 \) is the initial velocity, \( t \) is the time in seconds, and \( a \) is the acceleration. If an object starts from rest, like the box in our example, the initial velocity \( v_0 \) is 0. This simplifies our formula to:\[ d = \frac{1}{2} a t^2 \]Plugging in the given values, such as the acceleration of 2.0 m/s² and the time of 7.0 seconds, we calculate the displacement as 49.0 meters. This tells us that over the course of 7 seconds, with the given acceleration, the box travels 49 meters along the floor.Understanding how to calculate displacement ensures that we can precisely determine how far an object will move under specific conditions, which is crucial in applications ranging from engineering to everyday problem-solving.
Constant Acceleration Motion
Constant acceleration motion refers to scenarios where an object's acceleration does not change over time. This can occur in real-world situations, such as an object sliding down an inclined plane or a car accelerating at a steady rate on a straight path. A constant acceleration means that every second, the object's velocity changes uniformly.With a constant acceleration, we can use a set of kinematic equations to predict the future position and velocity of an object. The equation for displacement, \[ d = v_0 t + \frac{1}{2} a t^2 \], and the formula for force via Newton's second law, are crucial in these analyses.
Applying these concepts to our exercise, the acceleration was kept constant at 2.0 m/s² for a duration of 7.0 seconds. This allowed us to determine both the displacement and the force acting on the box as it moved. These steady conditions made it possible to accurately calculate the work done on the box, as the consistent acceleration led to a predictable displacement.
By mastering constant acceleration scenarios, students can better understand how forces influence motion, which is vital for designing systems and solving practical engineering problems.

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Most popular questions from this chapter

We usually neglect the mass of a spring if it is small compared to the mass attached to it. But in some applications, the mass of the spring must be taken into account. Consider a spring of unstretched length \(\ell\) and mass \(M_{\mathrm{S}}\) uniformly distributed along the length of the spring. A mass \(m\) is attached to the end of the spring. One end of the spring is fixed and the mass \(m\) is allowed to vibrate horizontally without friction (Fig. \(7-30\) ). Each point on the spring moves with a velocity proportional to the distance from that point to the fixed end. For example, if the mass on the end moves with speed \(v_{0}\), the midpoint of the spring moves with speed \(v_{0} / 2 .\) Show that the kinetic energy of the mass plus spring when the mass is moving with velocity \(v\) is $$ K=\frac{1}{2} M v^{2} $$ where \(M=m+\frac{1}{3} M_{\mathrm{S}}\) is the "effective mass" of the system. [Hint: Let \(D\) be the total length of the stretched spring. Then the velocity of a mass \(d m\) of a spring of length \(d x\) located at \(x\) is \(v(x)=v_{0}(x / D) .\) Note also that $$ d m=d x\left(M_{\mathrm{S}} / D\right) $$.

A constant force \(\overrightarrow{\mathbf{F}}=(2.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}}) \mathbf{N}\) acts on an object as it moves along a straight- line path. If the object's displacement is \(\overrightarrow{\mathbf{d}}=(1.0 \hat{\mathbf{i}}+5.0 \mathbf{j}) \mathrm{m},\) calculate the work done by \(\overrightarrow{\mathbf{F}}\) using these alternate ways of writing the dot product: (a) \(W=F d \cos \theta\) \(;\) (b) \(W=F_{x} d_{x}+F_{y} d_{y}\)

A 3.0 -m-long steel chain is stretched out along the top level of a horizontal scaffold at a construction site, in such a way that \(2.0 \mathrm{~m}\) of the chain remains on the top level and \(1.0 \mathrm{~m}\) hangs vertically, Fig. \(7-26 .\) At this point, the force on the hanging segment is sufficient to pull the entire chain over the edge. Once the chain is moving, the kinetic friction is so small that it can be neglected. How much work is performed on the chain by the force of gravity as the chain falls from the point where \(2.0 \mathrm{~m}\) remains on the scaffold to the point where the entire chain has left the scaffold? (Assume that the chain has a linear weight density of \(18 \mathrm{~N} / \mathrm{m}\).)

(I) Show that \(\overrightarrow{\mathbf{A}} \cdot(-\overrightarrow{\mathbf{B}})=-\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}\).

In the game of paintball, players use guns powered by pressurized gas to propel \(33 -\) g gel capsules filled with paint at the opposing team. Game rules dictate that a paintball cannot leave the barrel of a gun with a speed greater than 85\(\mathrm { m } / \mathrm { s } .\) Model the shot by assuming the pressurized gas applies a constant force \(F\) to a \(33 - g\) capsule over the length of the \(32 - \mathrm { cm }\) barrel. Determine \(F ( a )\) using the work-energy principle, and \(( b )\) using the kinematic equations and Newton's second law.

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