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Chapter 7: Problem 31

The only force acting on a \(2.0 \mathrm{~kg}\) moves along a positive \(x\) axis has an \(x\) component\(F_{x}=-6 x \mathrm{~N}\) with \(x\) in meters. The velocity at \(x=3.0 \mathrm{~m}\) is \(8.0 \mathrm{~m} / \mathrm{s} .\) (a) What is the velocity of the body at \(x=4.0 \mathrm{~m} ?\) (b) At what positive value of \(x\) will the body have a velocity of \(5.0 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
(a) The velocity at \(x=4.0\, \text{m}\) is approximately \(6.56\, \text{m/s}\). (b) The body will have a velocity of \(5.0\, \text{m/s}\) at \(x \approx 4.69\, \text{m}\).

Step by step solution

01

Identify the Given Information

We need to identify the given values in the problem:- Mass of the body: \(m = 2.0\, \text{kg}\).- Force component along the x-axis: \(F_{x} = -6x\, \text{N}\).- Initial velocity at \(x = 3.0\, \text{m}\) is \(v = 8.0\, \text{m/s}\).
02

Use Work-Energy Theorem

The work-energy theorem states that the change in kinetic energy is equal to the work done. First, we calculate the kinetic energy at the initial position \(x = 3\, \text{m}\):\[KE_1 = \frac{1}{2}mv^2 = \frac{1}{2} \cdot 2.0 \cdot (8.0)^2 = 64\, \text{J}\, \text{(joules)}\]Next, calculate the work done by the force \(F_x\) from \(x = 3\, \text{m}\) to \(x = 4\, \text{m}\):\[W = \int_{3}^{4} F_x \, dx = \int_{3}^{4} -6x \, dx = [-3x^2]_{3}^{4} = -3(4^2) + 3(3^2) = -48 + 27 = -21 \, \text{J}\]
03

Determine the Velocity at x = 4.0 m

The final kinetic energy at \(x = 4\, \text{m}\) can be calculated by adding work done to the initial total energy:\[KE_2 = KE_1 + W = 64 - 21 = 43\, \text{J}\]Thus, the velocity at \(x = 4.0\, \text{m}\) is given as:\[v^2 = \frac{2 \cdot KE_2}{m} = \frac{2 \cdot 43}{2} = 43 \quad \Rightarrow \quad v = \sqrt{43} \approx 6.56 \, \text{m/s}\]
04

Find Position where Velocity is 5.0 m/s

Use the kinetic energy expression when the velocity is \(5.0 \mathrm{~m/s}\):\[KE_3 = \frac{1}{2} m (5.0)^2 = \frac{1}{2} \cdot 2 \cdot 25 = 25\, \text{J} \]Now, calculate the work done from \(x = 3.0\, \text{m}\) to the position \(x = x_f\):\[W = KE_3 - KE_1 = 25 - 64 = -39\, \text{J}\]This work is represented by the integral from \(3\) meters to \(x_f\):\[-39 = \int_{3}^{x_f} -6x \, dx = [-3x^2]_{3}^{x_f} = -3x_f^2 + 27\]Solving for \(x_f\):\[-39 = -3x_f^2 + 27\]\[-66 = -3x_f^2\]\[x_f^2 = 22\]\[x_f = \sqrt{22} \approx 4.69\, \text{m}\]

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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 a vital concept in physics. It represents the energy of an object in motion. The faster an object moves, the more kinetic energy it possesses. Kinetic energy is mathematically defined as:
  • \( KE = \frac{1}{2}mv^2 \)
Here, \(m\) stands for mass and \(v\) for velocity. This formula shows that kinetic energy is proportional to the mass of the object and the square of its velocity. That's why even a slight increase in velocity results in a significant rise in kinetic energy.
Let's apply this understanding to our exercise: Given a mass of \(2.0\, \text{kg}\), and an initial velocity of \(8.0\, \text{m/s}\), the initial kinetic energy at \(x = 3.0\, \text{m}\) is calculated as \(64\, \text{J}\) (joules). Changes in velocity, as the force acts, will thus alter the kinetic energy, embodying the interplay between forces and motion.
Integral Calculus
Integral calculus is the branch of mathematics concerned with calculating the area under a curve. In physics, it helps us determine the total effect of a varying force over a distance. In this exercise, the force component \( F_{x} = -6x \) is a function of position \(x\), indicating it varies as the object moves along the x-axis.
To find the work done by this force in moving the object from \(x = 3\) meters to \(x = 4\) meters, we calculate the integral of the force over this interval:
  • \( W = \int_{3}^{4} -6x \; dx \)
This results in
  • \( [-3x^2]_{3}^{4} = -21 \; \text{J} \)
which shows the work needed due to the force acting over the specific distance. This work directly affects the kinetic energy of the object and helps us determine subsequent velocities.
Force Component
When dealing with problems involving motion and forces, it is crucial to understand force components, especially in directions of interest, like along the x-axis in our exercise.
For our given situation, the force component is \( F_{x} = -6x N \). This means the force changes with position, having a linear dependency on \(x\).
  • Negative sign shows the force opposes the motion in the positive x-direction.
This situation of a variable force is common in physics, and understanding it is key to dissecting motion behavior. This force model affects the body鈥檚 velocity by changing kinetic energy through work done, dictated by how the force behaves across the distance (in this case, decreasing quadratically).
Ultimately, calculations surrounding moving bodies rely on combining these variable forces to predict motion outcomes and understand energy transitions, enhancing our grasp of fundamental mechanics.

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Most popular questions from this chapter

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of \(12^{\circ}\) with the horizontal. The rope moves parallel to the slope with a constant speed of \(1.0 \mathrm{~m} / \mathrm{s}\). The force of the rope does \(900 \mathrm{~J}\) of work on the skier as the skier moves a distance of \(8.0 \mathrm{~m}\) up the incline. (a) If the rope moved with a constant speed of \(2.0 \mathrm{~m} / \mathrm{s}\), how much work would the force of the rope do on the skier as the skier moved a distance of \(8.0 \mathrm{~m}\) up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) \(1.0 \mathrm{~m} / \mathrm{s}\) and (c) \(2.0 \mathrm{~m} / \mathrm{s} ?\)

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 \(F(x)\).

A helicopter lifts a \(72 \mathrm{~kg}\) astronaut \(15 \mathrm{~m}\) vertically from the ocean by means of a cable. The acceleration of the astronaut is \(g / 10 .\) How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?

As a particle moves along an \(x\) axis, a force in the positive direction of the axis acts on it. Figure \(7-50\) shows the magnitude \(F\) of the force versus position \(x\) of the particle. The curve is given by \(F=a / x^{2}\), with \(a=9.0 \mathrm{~N} \cdot \mathrm{m}^{2} .\) Find the work done on the particle by the force as the particle moves from \(x=1.0 \mathrm{~m}\) to \(x=3.0 \mathrm{~m}\) by \((\mathrm{a})\) estimating the work from the graph and (b) integrating the force function.

An iceboat is at rest on a frictionless frozen lake when a sudden wind exerts a constant force of \(200 \mathrm{~N},\) toward the east, on the boat. Due to the angle of the sail, the wind causes the boat to slide in a straight line for a distance of \(8.0 \mathrm{~m}\) in a direction \(20^{\circ}\) north of east. What is the kinetic energy of the iceboat at the end of that \(8.0 \mathrm{~m} ?\)
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