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Chapter 6: Problem 17

How much work is required to move an object from \(x=0\) to \(x=3\) (measured in meters) in the presence of a force (in \(\mathrm{N}\) ) given by \(F(x)=2 x\) acting along the \(x\) -axis?

Short Answer

Expert verified
Answer: The work done on the object is 9 Joules.

Step by step solution

01

Write down the formula for work done

To calculate the work done when moving the object, we use the formula: Work = \(\int_{x_1}^{x_2} F(x) dx\) where \(F(x)\) is the force function and x1 and x2 are the initial and final positions of the object on the x-axis.
02

Identify the given Force function and the limits of integration

We are given the force function as: \(F(x) = 2x\) We are also given the initial and final positions of the object: Initial position (\(x_1\)): 0m Final position (\(x_2\)): 3m
03

Set up the integral for work done

Using the given information, we can set up the integral for work done: Work = \(\int_{0}^{3} 2x dx\)
04

Evaluate the integral

Now, evaluate the integral: Work = \(2\int_{0}^{3} x dx=\left[x^2\right]_0^3\)
05

Calculate work done

Plug in the limits of integration to get the numerical value for work done: Work = \((3^2) - (0^2) = 9 - 0 = 9\)
06

Interpret the result

The work required to move the object from x=0 to x=3 in the presence of the given force is 9 Joules.

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Most popular questions from this chapter

Calculate the work required to stretch the following springs 0.5 m from their equilibrium positions. Assume Hooke's law is obeyed. a. A spring that requires a force of \(50 \mathrm{N}\) to be stretched \(0.2 \mathrm{m}\) from its equilibrium position. b. A spring that requires \(50 \mathrm{J}\) of work to be stretched \(0.2 \mathrm{m}\) from its equilibrium position.

A rigid body with a mass of \(2 \mathrm{kg}\) moves along a line due to a force that produces a position function \(x(t)=4 t^{2},\) where \(x\) is measured in meters and \(t\) is measured in seconds. Find the work done during the first 5 s in two ways. a. Note that \(x^{\prime \prime}(t)=8 ;\) then use Newton's second law \(\left(F=m a=m x^{\prime \prime}(t)\right)\) to evaluate the work integral \(W=\int_{x_{0}}^{x_{f}} F(x) d x,\) where \(x_{0}\) and \(x_{f}\) are the initial and final positions, respectively. b. Change variables in the work integral and integrate with respect to \(t .\) Be sure your answer agrees with part (a).

Compute the following derivatives using the method of your choice. $$\frac{d}{d x}\left(1+\frac{4}{x}\right)^{x}$$

Archimedes' principle says that the buoyant force exerted on an object that is (partially or totally) submerged in water is equal to the weight of the water displaced by the object (see figure). Let \(\rho_{w}=1 \mathrm{g} / \mathrm{cm}^{3}=1000 \mathrm{kg} / \mathrm{m}^{3}\) be the density of water and let \(\rho\) be the density of an object in water. Let \(f=\rho / \rho_{w}\). If \(01,\) then the object sinks. Consider a cubical box with sides 2 m long floating in water with one-half of its volume submerged \(\left(\rho=\rho_{w} / 2\right) .\) Find the force required to fully submerge the box (so its top surface is at the water level).

Carry out the following steps to derive the formula \(\int \operatorname{csch} x d x=\ln |\tanh (x / 2)|+C\) (Theorem 9). a. Change variables with the substitution \(u=x / 2\) to show that $$\int \operatorname{csch} x d x=\int \frac{2 d u}{\sinh 2 u}$$. b. Use the identity for \(\sinh 2 u\) to show that \(\frac{2}{\sinh 2 u}=\frac{\operatorname{sech}^{2} u}{\tanh u}\). c. Change variables again to determine \(\int \frac{\operatorname{sech}^{2} u}{\tanh u} d u,\) and then express your answer in terms of \(x\).
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