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Chapter 6: Problem 88

Suppose that the resistive force of the air on a skydiver can be approximated by \(f=-b v^{2} .\) If the terminal velocity of a 50.0 -kg skydiver is \(60.0 \mathrm{m} / \mathrm{s}\), what is the value of \(b ?\)

Short Answer

Expert verified
The value of \(b\) is 0.1367 \(kg/m\).

Step by step solution

01

Understand the situation at terminal velocity

At terminal velocity, the skydiver is no longer accelerating. Thus, the total force acting on the skydiver must be zero. Two forces are acting on the skydiver: the gravitational force and the resistive force. Therefore, Gravity force = Resistive force, i.e., \(mg = bv^2\).
02

Apply the values to find the resistive constant

Now replace \(m\), \(g\), and \(v\) with actual values given in the problem to solve for \(b\). The gravitational constant, \(g\), is 9.81 \(m/s^2\). So \(b = \frac{mg}{v^2} = \frac{50kg * 9.81 m/s^2}{(60m/s)^2}\)
03

Calculate the value of b

Now calculate the value of \(b\). After solving the above equation, the calculated value of \(b\) is 0.1367 \(kg/m\).

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Key Concepts

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

Resistive Force
The resistive force is an essential concept to understand when discussing objects in motion through a fluid, like air. In our case, this force is acted upon by air Resistance, which slows down a skydiver. This force is mathematically represented as \( f = -b v^2 \). Here,
  • \(f\) is the resistive force.
  • \(b\) is the resistive constant.
  • \(v\) is the velocity of the object.
The negative sign indicates that the force acts in the opposite direction to the velocity. As the velocity increases, the resistive force, acting against the motion, also increases. This eventually leads to a state where the object reaches a terminal velocity where no further acceleration occurs.
Gravitational Force
Gravitational force plays a vital role in the journey of a skydiver. It is the force due to gravity that pulls objects toward the Earth's center. The gravitational force acting on an object of mass \(m\) near the Earth鈥檚 surface is given by the product of mass \(m\) and the gravitational acceleration \(g\), written as \( F_g = mg \). Here,
  • \(F_g\) is the gravitational force.
  • \(m\) is the mass of the skydiver.
  • \(g\) is the acceleration due to gravity, approximately \(9.81 m/s^2\).
In the context of terminal velocity, the gravitational force must equal the resistive force for the skydiver to stop accelerating. This balance is what ultimately brings the skydiver to a steady terminal velocity.
Resistive Constant
The resistive constant, denoted by \(b\), is a crucial parameter in the equation of resistive force, \(f = -b v^2\). It quantifies the extent of resistance a fluid, like air, imposes on an object moving through it. In the given problem, the value of \(b\) was calculated using the equation at terminal velocity, where all acting forces balance out, such that the gravitational force equals the resistive force. Given that \(mg = bv^2\), it turns out that
  • \(b\) can be expressed as \(b = \frac{mg}{v^2}\).
  • Substituting the mass (50 kg), gravitational acceleration (9.81 m/s虏), and terminal velocity (60 m/s), we find \(b = 0.1367\) kg/m.
The resistive constant, thus, depends on these factors and is specific to the conditions of each unique scenario. It provides insights into how changes in velocity or shape may influence the resistance experienced by the skydiver.

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Most popular questions from this chapter

To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at \(g\). Why do they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?

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