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Chapter 6: Problem 27

A steady force of \(500 \mathrm{~N}\) is applied horizontally to push a loaded cart at a constant speed. How far would the cart move when 3500 J of work is done on it by that applied force?

Short Answer

Expert verified
The cart moves 7 meters.

Step by step solution

01

Understand the Problem

You have a force of \(500\,\text{N}\) applied horizontally on a cart and need to calculate how far the cart moves when \(3500\,\text{J}\) of work is done.
02

Recall the Work Formula

The formula for work when a constant force is applied in the direction of motion is \( W = F \cdot d \), where \( W \) is the work done, \( F \) is the force applied, and \( d \) is the distance moved.
03

Solve for Distance

Rearrange the work formula to solve for distance: \( d = \frac{W}{F} \).
04

Substitute Known Values

Substitute the values for work \( W = 3500\,\text{J} \) and force \( F = 500\,\text{N} \) into the formula: \( d = \frac{3500}{500} \).
05

Calculate the Distance

Perform the division: \( d = 7 \). Therefore, the cart moves \(7\,\text{m}\).

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Key Concepts

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

Work Formula
In the realm of physics, understanding how force and motion interact is essential. This is where the concept of work comes into play. The mathematical expression for work when dealing with a constant force in the direction of motion is given by the formula: \( W = F \cdot d \). Here, \( W \) represents the work done in joules, \( F \) is the constant force applied in newtons, and \( d \) is the distance the object travels in meters.

The work formula simplifies the calculation of energy transfer through movement. This equation shows us how energy is transferred from one system (exerting the force) to another (the object being moved). This energy transfer occurs through mechanical work, which depends on both the magnitude of the force applied and the distance over which the force is exerted.

Understanding this fundamental equation helps in numerous real-world scenarios. Whether it's moving a cart, lifting weights, or pushing a box, the work formula lays the foundation for analyzing how much energy is needed to perform different tasks.
Constant Force
A crucial part of calculating work is the idea of constant force. A force is considered constant if its magnitude and direction remain steady over time as it acts on an object. This constant force makes it easier to predict how objects will move and interact.

In the given problem, a steady force of \(500\,\text{N}\) is applied consistently in one direction. This implies that the force doesn't vary while pushing the cart. Because the force is constant, the work formula \( W = F \cdot d \) can be applied effectively.
  • Constant force is generally easier to analyze mathematically.
  • It eliminates variables due to changes in the force over time.
  • Allows for straightforward calculations for work and energy.
Without a constant force, the calculations can become more complex, requiring calculus to account for changes in force over time. Hence, constant force scenarios are essential to building foundational skills in physics.
Distance Calculation
Once you understand the work formula and the presence of a constant force, the next step is to determine how far an object will move. To calculate the distance when work and force are known, you rearrange the work formula to \( d = \frac{W}{F} \).

This rearranged formula highlights a direct relationship between work, force, and distance. When you substitute the known values of work \( (3500 \text{ J}) \) and force \( (500 \text{ N}) \) into this formula, you get:
  • \( d = \frac{3500}{500} \)
  • \( d = 7 \)
Therefore, the cart travels a total distance of \(7\,\text{m}\). This calculation demonstrates how energy, through work, directly influences the distance an object moves, given a constant force. By understanding how to calculate this distance, you can better grasp the practical implications of energy transfer and mechanical movement in everyday situations.

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Most popular questions from this chapter

In Fig. 6-1, assume that the object is being pulled in a straight line along the ground by a 75-N force directed \(28^{\circ}\) above the horizontal. How much work does the force do in pulling the object \(8.0 \mathrm{~m}\) horizontally? The work done is equal to the product of the displacement, \(8.0 \mathrm{~m}\), and the component of the force that is parallel to the displacement, \((75 \mathrm{~N})\left(\cos 28^{\circ}\right)\). Thus, $$ W=(75 \mathrm{~N})\left(\cos 28^{\circ}\right)(8.0 \mathrm{~m})=0.53 \mathrm{~kJ} $$

A 2000 -kg elevator rises from rest in the basement to the fourth floor, a distance of \(25 \mathrm{~m}\). As it passes the fourth floor, its speed is \(3.0 \mathrm{~m} / \mathrm{s}\). There is a constant frictional force of \(500 \mathrm{~N}\). Calculate the work done by the lifting mechanism.

An automobile is pushed \(10.0 \mathrm{ft}\) by a woman exerting \(80.0 \mathrm{lb}\) of force horizontally on the vehicle. How much work does she do (a) in \(\mathrm{ft} \cdot \mathrm{lb}\) and \((b)\) in joules?

A 60 000-kg train is being dragged along a straight line up a \(1.0\) percent grade (i.e., the road rises \(1.0 \mathrm{~m}\) for each \(100 \mathrm{~m}\) traveled horizontally) by a steady drawbar pull of \(3.0 \mathrm{kN}\) parallel to the incline. The friction force opposing the motion of the train is \(4.0\) \(\mathrm{kN}\). The train's initial speed is \(12 \mathrm{~m} / \mathrm{s}\). Through what distance s will the train move along its tracks before its speed is reduced to \(9.0 \mathrm{~m} / \mathrm{s}\) ? Use energy considerations. The change in total energy of the train is due to the work done by the friction force (which is negative) and the drawbar pull (which is positive): Change in \(\mathrm{KE}+\) change in \(\mathrm{PE}_{\mathrm{G}}=W_{\text {drawbar }}+W_{\text {friction }}\) The train loses \(\mathrm{KE}\) and gains \(\mathrm{PE}_{\mathrm{G}}\). It rises a height \(h=s \sin \theta\), where \(\theta\) is the incline angle and \(\tan \theta=1 / 100\). Hence, \(\theta=0.573^{\circ}\), and \(h=0.010 \mathrm{~s}\) (at small angles \(\tan \theta \approx \sin \theta\) ). Therefore, $$ \begin{array}{c} \frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)+m g(0.010 s)=(3000 \mathrm{~N})(s)(1)+(4000 \mathrm{~N})(s)(-1) \\ -1.89 \times 10^{6} \mathrm{~J}+\left(5.89 \times 10^{3} \mathrm{~N}\right) s=(-1000 \mathrm{~N}) s \end{array} $$ from which we obtain \(s=274 \mathrm{~m}=0.27 \mathrm{~km}\).

Calculate the average power required to raise a 150 -kg drum to a height of \(20 \mathrm{~m}\) in a time of \(1.0\) minute. Give your answer in both kilowatts and horsepower.
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