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Chapter 6: Problem 45

A force in the \(+x\) -direction with magnitude \(F(x)=18.0 \mathrm{~N}-(0.530 \mathrm{~N} / \mathrm{m}) x\) is applied to a \(6.00 \mathrm{~kg}\) box that is sitting on the horizontal, frictionless surface of a frozen lake. \(F(x)\) is the only horizontal force on the box. If the box is initially at rest at \(x=0\) what is its speed after it has traveled \(14.0 \mathrm{~m} ?\)

Short Answer

Expert verified
To find the final speed, first calculate the work done by the varying force as the box moves from 0 to 14.0 m using integral calculus, and then equate this work to the kinetic energy using the work-energy theorem to get the final velocity.

Step by step solution

01

Applying the work-energy theorem

The work done (W) on the box is equal to change in kinetic energy (鈭咾). Since the box is initially at rest, its initial kinetic energy is zero. Hence, the work done is simply equal to its final kinetic energy. So, our goal is to compute the total work done by the force which is the integral of force with respect to displacement from initial to final position, and equate it to the final kinetic energy. Mathematically, it is \[W = \Delta K = \frac{1}{2} m v_f^2\] where \(v_f\) is the final velocity of the box.
02

Calculating the work done

The total work done by the force is given by integrating the force function from the initial position (0) to the final position (14.0 m). Using the given force function F(x) = 18.0 N - 0.530 N/m * x, the work done is: \[W = \int_0^{14} F(x) dx = \int_0^{14} (18 - 0.530x) dx\] Evaluating this definite integral results in the total work done on the box in Joules.
03

Equating Work and Kinetic Energy

Now that we calculated W, we can calculate the final speed. We have kinetic energy as: \[\Delta K = W = \frac{1}{2} m v_f^2\], where m is the mass of the box (6 kg) and \(v_f\) is the final speed. Solve the equation for the final speed \(v_f\).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It depends on both the mass of the object and its velocity. The formula for kinetic energy is:
  • \[ K = \frac{1}{2} mv^2 \]
where \(K\) is the kinetic energy, \(m\) is the mass, and \(v\) is the velocity of the object.
In our exercise, the box starts at rest, so its initial kinetic energy is zero. This means that any work done on the box will directly change its kinetic energy, giving it a velocity at the end of the process. When the force is applied, the box's velocity changes, increasing its kinetic energy from zero to some final value that we need to find.
Force Integration
Force integration is a method used to calculate the work done by a variable force over a certain displacement. In our problem, the force acting on the box is not constant; it decreases with displacement \(x\) due to the term \(-0.530 \mathrm{~N/m} \times x\).
To find the work done, we integrate the given force function over the interval of motion:
  • \[ W = \int_0^{14} (18.0 - 0.530x) \, dx \]
This integral sums up the small amounts of work done over each infinitesimal segment \(dx\) of the path from 0 to 14 meters. The integral calculation results in total work done, which represents the area under the force vs. displacement curve.
Work Done Calculation
Calculating work done refers to finding the total energy transferred to the object through the force applied over a distance. Using the equation for work done from our force function:
  • \[ W = \int_0^{14} (18.0 - 0.530x) \, dx \]
we evaluate this integral to determine how much energy the force has imparted to the box as it moved.
Performing the integration gives:
  • \[ W = \left[ 18.0x - \frac{0.530}{2}x^2 \right]_0^{14} \]
  • Calculating this expression, the result for \(W\) will be in joules, as it's the standard unit for work and energy in physics.
This value then represents the complete energy transferred to the box while moving across the frozen lake.
Final Speed Determination
To find the final speed of the box, we use the work-energy theorem, which relates work done to the change in kinetic energy:
  • \[ \Delta K = W = \frac{1}{2} m v_f^2 \]
Given the box's mass \(m = 6 \mathrm{~kg}\), and knowing \(W\) from the work done calculation, we can solve for the final velocity \(v_f\). Rearrange the kinetic energy equation to:
  • \[ v_f^2 = \frac{2W}{m} \]
  • \[ v_f = \sqrt{\frac{2W}{m}} \]
By plugging in the values for \(W\) and \(m\), we can determine how fast the box is moving after traveling 14 meters. This result reaffirms the relationship between force, work done, and kinetic energy, showing how motion is influenced by applied forces.

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Most popular questions from this chapter

A net force along the \(x\) -axis that has \(x\) -component \(F_{x}=-12.0 \mathrm{~N}+\left(0.300 \mathrm{~N} / \mathrm{m}^{2}\right) x^{2}\) is applied to a \(5.00 \mathrm{~kg}\) object that is initially at the origin and moving in the \(-x\) -direction with a speed of \(6.00 \mathrm{~m} / \mathrm{s}\). What is the speed of the object when it reaches the point \(x=5.00 \mathrm{~m} ?\)

A \(1.50 \mathrm{~kg}\) book is sliding along a rough horizontal surface. cats, have a mass of about \(70 \mathrm{~kg}\) and have been clocked to run at up to \(72 \mathrm{mi} / \mathrm{h}(32 \mathrm{~m} / \mathrm{s}) .\) (a) How many joules of kinetic energy does such a swift cheetah have? (b) By what factor would its kinetic energy change \text { if its speed were doubled? }

You push your physics book \(1.50 \mathrm{~m}\) along a horizontal tabletop with a horizontal push of \(2.40 \mathrm{~N}\) while the opposing force of friction is \(0.600 \mathrm{~N}\). How much work does each of the following forces do on the book: (a) your \(2.40 \mathrm{~N}\) push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?

A ski tow operates on a \(15.0^{\circ}\) slope of length \(300 \mathrm{~m}\). The rope moves at \(12.0 \mathrm{~km} / \mathrm{h}\) and provides power for 50 riders at one time, with an average mass per rider of \(70.0 \mathrm{~kg}\). Estimate the power required to operate the tow.

Object \(A\) has \(27 \mathrm{~J}\) of kinetic energy. Object \(B\) has one-quarter the mass of object \(A\). (a) If object \(B\) also has \(27 \mathrm{~J}\) of kinetic energy, is it moving faster or slower than object \(A\) ? By what factor? (b) By what factor does the speed of each object change if total work \(-18 \mathrm{~J}\) is done on each?
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