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Chapter 6: Problem 70

CALC 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 \(x=5.00 \mathrm{m} ?\)

Short Answer

Expert verified
The speed is approximately 4.12 m/s.

Step by step solution

01

Analyze the Force Expression

The force acting along the x-axis is given by \( F_x = -12.0 \, \text{N} + (0.300 \, \text{N/m}^2) x^2 \). This is a function of position \( x \), which means that the force varies as the object moves along the x-axis.
02

Determine Work Done by the Force

The work done by the force when the object moves from an initial position \( x_0 = 0 \) to a final position \( x = 5.00 \, \text{m} \) is given by the definite integral \( W = \int_{0}^{5.00} F_x \, dx \). This integral will help us find the change in kinetic energy.
03

Calculate the Definite Integral

Compute the integral \( W = \int_{0}^{5.00} \left(-12.0 + 0.300x^2\right) \, dx \):\[W = \left[-12.0x + \frac{0.300}{3}x^3\right]_{0}^{5.00} = \left[-12.0(5.00) + 0.1(5.00)^3 \right] - \left[0\right]\]\[W = (-60.0 + 12.5) \, \text{J} = -47.5 \, \text{J}\]This means 47.5 J of work is done on the object.
04

Apply the Work-Energy Theorem

According to the work-energy theorem, the work done by the net force is equal to the change in kinetic energy of the object, \( W = \Delta KE = KE_f - KE_i \).Let \( v_f \) be the final velocity of the object. The initial kinetic energy, \( KE_i = \frac{1}{2}mv_i^2 \) and the final kinetic energy \( KE_f = \frac{1}{2}mv_f^2 \).
05

Solve for the Final Velocity

Substitute the known values and solve for \( v_f \):\[ -47.5 = \frac{1}{2}(5.00)v_f^2 - \frac{1}{2}(5.00)(-6.00)^2 \]\[ -47.5 = \frac{1}{2}(5.00)v_f^2 - 90.0 \]\[ 42.5 = \frac{1}{2}(5.00)v_f^2 \]\[ 85.0 = 5.00v_f^2 \]\[ v_f^2 = 17.0 \]\[ v_f = \sqrt{17.0} \approx 4.12 \, \text{m/s} \]The final speed of the object is approximately \( 4.12 \, \text{m/s} \).

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Key Concepts

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

Net Force
The concept of net force is crucial when analyzing how an object moves under the influence of multiple forces. Net force is essentially the vector sum of all the individual forces acting on an object. It determines whether the object's motion changes, and by how much. In our specific problem, the net force along the x-axis is described by a function of position:
  • \( F_x = -12.0 \, \text{N} + (0.300 \, \text{N/m}^2) x^2 \)

This force is not constant; it changes with the object's position along the x-axis. As the object moves, this function accounts for how much force is exerted on the object at each point along its path. This variable force is critical in calculating the work done, as it affects the object's overall energy and ultimately, its velocity. Understanding net force is the key to using the work-energy theorem effectively.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It's given by the expression:
  • \( KE = \frac{1}{2}mv^2 \)
where \(m\) is the mass of the object and \(v\) is its velocity. In this problem, kinetic energy plays a pivotal role as we use the work-energy theorem to find the final speed of the object. Initially, the object is moving with a velocity of -6.00 m/s, so its initial kinetic energy can be calculated as:
  • \( KE_i = \frac{1}{2}(5.00)(-6.00)^2 \)

As the object moves and work is done on it by the net force, its kinetic energy changes. This change is given by:
  • \( \Delta KE = KE_f - KE_i \)

where \( KE_f \) is the final kinetic energy. By calculating the work done and knowing the initial kinetic energy, we can isolate the final velocity, \(v_f\), demonstrating how energy transformations govern motion.
Definite Integral
In physics, a definite integral is often used to calculate the total effect of a varying quantity over a certain range. In this exercise, the definite integral helps us find the total work done by a position-dependent force as the object moves from one point to another:
  • \( W = \int_{0}^{5.00} F_x \, dx \)
Calculating this integral with the given force function, \( F_x = -12.0 + 0.300x^2 \), accounts for how the force acts over each small segment of the object's journey. The resulting work obtained from
  • \( W = -47.5 \, \text{J} \)
indicates how much energy is transferred to or from the object due to this force over the 5-meter path. Working through definite integrals is a practical application of calculus that enables one to solve real-world physics problems involving forces that change with position.

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Most popular questions from this chapter

CALC The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the earth's surface this force is equal to the object's normal weight \(m g,\) where \(g=9.8 \mathrm{m} / \mathrm{s}^{2},\) and at large distances, the force is zero. If a \(20,000-\mathrm{kg}\) asteroid falls to earth from a very great distance away, what will be its minimum speed as it strikes the earth's surface, and how much kinetic energy will it impart to our planet? You can ignore the effects of the earth's atmosphere.

CALC An airplane in flight is subject to an air resistance force proportional to the square of its speed \(v .\) But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward (Fig. P6.104). The upward force is the lift force that keeps the airplane aloft, and the backward force is called induced drag. At flying speeds, induced drag is inversely proportional to \(v^{2},\) so that the total air resistance force can be expressed by \(F_{\text { air }}=\alpha v^{2}+\beta / v^{2},\) where \(\alpha\) and \(\beta\) are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna \(150,\) a small single-engine airplane, \(\alpha=0.30 \mathrm{N} \cdot \mathrm{s}^{2} / \mathrm{m}^{2}\) and \(\beta=3.5 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} .\) In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed (in \(\mathrm{km} / \mathrm{h} )\) at which this airplane will have the maximum range (that is, travel the greatest distance) for a given quantity of fuel. (b) Calculate the speed (in \(\mathrm{km} / \mathrm{h} )\) for which the airplane will have the maximum endurance (that is, remain in the air the longest time).

A sled with mass 8.00 \(\mathrm{kg}\) moves in a straight line on a frictionless horizontal surface. At one point in its path, its speed is \(4.00 \mathrm{m} / \mathrm{s} ;\) after it has traveled 2.50 \(\mathrm{m}\) beyond this point, its speed is 6.00 \(\mathrm{m} / \mathrm{s}\) . Use the work-energy theorem to find the force acting on the sled, assuming that this force is constant and that it acts in the direction of the sled's motion.

When its \(75-\mathrm{kW}(100 \mathrm{-hp})\) engine is generating full power, a small single-engine airplane with mass 700 \(\mathrm{kg}\) gains altitude at a rate of 2.5 \(\mathrm{m} / \mathrm{s}(150 \mathrm{m} / \mathrm{min}\) , or 500 \(\mathrm{ft} / \mathrm{min}\) ). What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 \(\mathrm{W}\) . The remainder of the day she is sitting in class, studying, or resting; during these activities, she expends energy at an average rate of 100 \(\mathrm{W}\) . If she expends a total of \(1.1 \times 10^{7} \mathrm{J}\) of energy in a 24 -hour day, how much of the day did she spend walking?
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