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A \(65.0-\mathrm{kg}\) skydiver reaches a terminal speed of \(55.0 \mathrm{~m} / \mathrm{s}\) with her parachute undeployed. Suppose the drag force acting on her is proportional to the speed squared, or \(F_{\text {dray }}=k v^{2}\). (a) What is the constant of proportionality k? (Assume the gravitational acceleration is \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).) (b) What was the magnitude of her acceleration when she was falling at half terminal speed?

Short Answer

Expert verified
The constant of proportionality 'k' is calculated to be approximately \(0.0342 kg/m\). The acceleration is calculated when the velocity is half the terminal velocity, which yields approximately \(4.9 m/s^2\).

Step by step solution

01

Initializing Given Values

Initialize given values: mass \(m = 65.0 kg\), terminal velocity \(v_t = 55.0 m/s\), and gravitational acceleration \(g = 9.80 m/s^2\).
02

Calculate Drag Force at Terminal Velocity

At terminal velocity, the net force on the skydiver is zero. So, the force due to gravity \(mg\) is equal to the drag force \(kv_t^2\). This gives us the equation: \(mg = kv_t^2\).
03

Solve for Constant 'k'

Radio the terms in the equation obtained in Step 2 to find 'k', which gives \(k = mg / v_t^2\). Substituting the given values, we find \(k = (65.0 kg * 9.8 m/s^2) / (55.0 m/s)^2\).
04

Calculate Constant 'k'

Evaluating the expression from Step 3 gives value of 'k'. This value will be used for part (b).
05

Understanding Part (b)

For the second part, we need to find the acceleration when the velocity is half of the terminal velocity (\(v_h = 0.5 * v_t\)), and we use the formula that acceleration \(a = net force / mass\). Let's find the net force.
06

Calculate Net Force and Acceleration

The net force when the velocity is half the terminal velocity is the difference between the force due to gravity (\(mg\)) and the drag force at that velocity (\(k*v_h^2\)). This gives us \(net force = mg - k*v_h^2\). We can find acceleration (\(a\)) now with formula \(a = net force / m\). Substituting the given values, we find the acceleration.

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Key Concepts

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

Drag Force
Imagine you're skydiving, and as you accelerate towards the Earth, you can feel the air resisting your movement. This resistance is known as drag force, and it plays a crucial role in determining the speed at which objects fall through a fluid (like air). In physics, drag force is often discussed in the context of objects reaching what is called terminal velocity, the constant speed where the force of gravity is balanced by the drag force, resulting in no further acceleration.

The drag force can be affected by a number of factors including the object’s speed, the density of the fluid, the cross-sectional area of the object, and the object’s shape. Mathematically, the drag force is commonly represented as being proportional to the square of the object’s velocity. The formula for calculating drag force at high speeds, where the force depends on the square of the speed, is given by: \( F_{drag} = k v^2 \) where \( F_{drag} \) is the drag force, \( v \) is the velocity of the object, and \( k \) is the constant of proportionality, which depends on factors such as the object's shape and the properties of the fluid.

Understanding the drag force is essential not only in skydiving but in various fields such as automotive design, aerospace engineering, and sports science, where reducing the drag can lead to improved performance and efficiency.
Constant of Proportionality
When dealing with equations in physics, the constant of proportionality is the factor that relates two variables in proportion to one another. In the context of the drag force, the constant of proportionality, designated by the letter \( k \), connects the drag force to the square of the velocity (speed).

In the case of the skydiver from our problem, the constant of proportionality \( k \) is calculated using the equation where the gravitational force (weight of the skydiver) is balanced with the drag force at terminal velocity: \( mg = kv_t^2 \). By rearranging the equation, we find \( k \) to be: \( k = \frac{mg}{v_t^2} \), where \( m \) is the mass of the skydiver, \( g \) is the acceleration due to gravity, and \( v_t \) is the terminal velocity.

The constant of proportionality is vital because it condenses all the factors impacting drag—like fluid density, object shape, and size—into one comparative value. By knowing this constant, we can predict the drag force experienced by the object at different velocities and thus calculate necessary quantities for various practical applications, such as the descent rate of parachutes or the fuel efficiency of vehicles.
Acceleration Due to Gravity
One of the most fundamental concepts in physics is the acceleration due to gravity, often denoted by \( g \). On Earth, this value is approximately \( 9.81 \text{m/s}^2 \), and it represents the rate at which objects accelerate towards the ground when falling.

The acceleration due to gravity is crucial for solving problems involving falling objects, like the skydiving problem. It helps us determine the force of gravity on the skydiver: \( F_{gravity} = mg \), where \( m \) is the mass. It's important to note that this acceleration assumes that the only force acting on the object is gravity, with no other forces like air resistance or drag.

In scenarios where other forces are present, the net acceleration of the object will differ from \( g \). For the skydiver at half the terminal speed, we take into account the reduced drag force along with the constant force of gravity to find the actual acceleration. This gives us a practical understanding of how objects behave under the influence of gravity and other forces, providing insights for everything from designing safety measures for high-speed sports to enhancing the realism in computer simulations.

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Most popular questions from this chapter

A high diver of mass \(70.0 \mathrm{~kg}\) jumps off a board \(10.0 \mathrm{~m}\) above the water. If her downward motion is stopped \(2.00 \mathrm{~s}\) after she enters the water, what average upward force did the water exert on her?

A coin is placed near one edge of a book lying on a table, and that edge of the book is lifted until the coin just slips down the incline as shown in Figure \(\mathrm{P} 4,70\). The angle of the incline, \(\theta_{c}\), called the critical angle, is measured. (a) Draw a frec-body diagram for the coin when it is on the verge of slipping and identify all forces acting on it. Your free- body diagram should include a force of static friction acting up the incline. (b) Is the magnitude of the friction force equal to \(\mu_{3} n\) for angles less than \(\theta_{c}\) ? Explain. What can you definitely say about the magnitude of the friction force for any angle \(\theta \leq \theta_{c} ?\) (c) Show that the coefficient of static friction is given by \(\mu_{\mathrm{s}}=\tan \theta_{c}\). (d) Once the coin starts to slide down the incline, the angle can be adjusted to a new value \(\theta_{c}^{\prime} \leq \theta_{c}\) such that the coins moves down the incline with constant speed. How does observation enable you to obtain the coefficient of kinetic friction?

A car is traveling at \(50.0 \mathrm{~km} / \mathrm{h}\) on a flat highway. (a) If the coefficient of friction between road and tires on a rainy day is \(0.100\), what is the minimum distance in which the car will stop? (b) What is the stopping distance when the surface is dry and the coefficient of friction is \(0.600 ?\)

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the \(\mathrm{cab}\), if the truck stops suddenly in an accident or even in braking. Assume, for example, a \(10000-\mathrm{kg}\) load sits on the flatbed of a 20 \(000-\mathrm{kg}\) truck moving at \(12.0 \mathrm{~m} / \mathrm{s}\). Assume the load is not tied down to the truck and has a coefficient of static friction of \(0.500\) with the truck bed. (a) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck. (b) Is any piece of data unnecessary for the solution?

A magician pulls a tablecloth from under a \(200-\mathrm{g} \mathrm{mug}\) located \(30.0 \mathrm{~cm}\) from the edge of the cloth. The cloth exerts a friction force of \(0.100 \mathrm{~N}\) on the mug and is pulled with a constant acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\). How far does the mug move relative to the horizontal tabletop before the cloth is completely out from under it? Note that the cloth must move more than \(30 \mathrm{~cm}\) relative to the tabletop during the process.

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