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\(\bullet\) \(\bullet\) Human terminal velocity. By landing properly and on soft ground (and by being lucky!), humans have survived falls from airplanes when, for example, a parachute failed to open, with astonishingly little injury. Without a parachute, a typical human eventually reaches a terminal velocity of about 62 \(\mathrm{m} / \mathrm{s}\) . Sup- pose the fall is from an airplane 1000 \(\mathrm{m}\) high. (a) How fast would a person be falling when he reached the ground if there were no air drag? (b) If a 70 kg person reaches the ground trav- eling at the terminal velocity of 62 \(\mathrm{m} / \mathrm{s}\) , how much mechanical energy was lost during the fall? What happened to that energy?

Short Answer

Expert verified
(a) 140 m/s without air drag. (b) 551,460 J lost as heat and sound.

Step by step solution

01

Calculate Final Velocity Without Air Resistance

Using the kinematic equation for motion without air resistance, where the initial velocity is 0 and acceleration is due to gravity \( g = 9.8 \ \mathrm{m/s^2} \), we can calculate the velocity just before hitting the ground. The equation is: \[ v = \sqrt{2gh} \]Plug in the height \( h = 1000 \ \mathrm{m} \):\[ v = \sqrt{2 \times 9.8 \times 1000} = \sqrt{19600} = 140 \ \mathrm{m/s} \]So, the final velocity without air resistance would be \( 140 \ \mathrm{m/s} \).
02

Identify the Terminal Velocity

Given in the problem, the terminal velocity of the human is \( 62 \ \mathrm{m/s} \). This is the velocity achieved when air resistance balances the weight of the person, causing the net force to be zero and the acceleration to cease.
03

Calculate Initial Mechanical Energy

We need to calculate the initial mechanical energy at the top of the fall. The mechanical energy at the top is purely potential energy, given by:\[ E_{\text{initial}} = mgh = 70 \times 9.8 \times 1000 = 686,000 \ \mathrm{J} \]
04

Calculate Final Mechanical Energy

At terminal velocity, the final mechanical energy is purely kinetic because there is negligible potential energy left at the ground. The kinetic energy is:\[ E_{\text{final}} = \frac{1}{2}mv^2 = \frac{1}{2} \times 70 \times 62^2 = \frac{1}{2} \times 70 \times 3844 = 134,540 \ \mathrm{J} \]
05

Determine Mechanical Energy Loss

The loss in mechanical energy is the difference between the initial potential energy and the final kinetic energy:\[ \Delta E = E_{\text{initial}} - E_{\text{final}} = 686,000 - 134,540 = 551,460 \ \mathrm{J} \]
06

Explain the Energy Loss

The mechanical energy lost, which is \( 551,460 \ \mathrm{J} \), was dissipated primarily as heat due to air friction and also slightly as sound when moving through the air.

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Key Concepts

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

Kinematic Equations
Kinematic equations are essential tools for understanding motion. They help us predict how objects behave under constant acceleration. By knowing some initial conditions, like starting speed and the rate of acceleration due to gravity, we can figure out things like final speed or distance traveled.

For instance, when ignoring air resistance, you can use a kinematic equation to calculate the speed of a falling object just before it hits the ground. The equation, \( v = \sqrt{2gh} \), tells us the final velocity \( v \) of an object falling from a height \( h \) with gravity \( g = 9.8 \ \mathrm{m/s^2} \) being the only force. This shows that, without air drag, a person falling from 1000 meters reaches a speed of 140 m/s.
  • This represents motion under ideal conditions.
  • It helps understand how gravity affects motion.
  • It simplifies calculations by ignoring friction and resistance.
Mechanical Energy
Mechanical energy is the sum of potential and kinetic energy in a system. It helps us analyze how energy is transformed and conserved during motion.

At the start of a fall, potential energy is at its maximum because of the height above the ground. This energy can be calculated using \( E_{\text{initial}} = mgh \). For a 70 kg person at 1000 meters, this is 686,000 Joules.

When the person hits the ground at terminal velocity, most of that energy has transformed into kinetic energy, calculated as \( E_{\text{final}} = \frac{1}{2}mv^2 \), which equals 134,540 Joules for a velocity of 62 m/s.
  • The difference in energy indicates an energy loss.
  • Initial energy is converted to both kinetic energy and energy lost to the environment.
  • Understanding energy conversion helps explain how systems behave and why energy seems to "disappear."
Air Resistance
Air resistance plays a significant role in real-world motion by opposing the motion of falling objects. It acts in the opposite direction to gravity and reduces acceleration until it becomes zero, resulting in terminal velocity.

When air resistance equals the gravitational force, the net force becomes zero, and the object falls at a constant speed known as terminal velocity. In this exercise, terminal velocity for a human was calculated or given as 62 m/s.

Here’s why air resistance is important:
  • It acts as a safety feature, like a natural parachute.
  • It dissipates mechanical energy as heat and sound, leading to a lower speed on impact.
  • It changes our idealized calculations of motion because no real system is without resistance.
Understanding air resistance helps in designing safety features and explaining why objects don't fall as fast as theoretically possible. It highlights the practical differences between theory and real-world applications.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of \(36.9^{\circ}\) below the horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction.

\(\bullet\) \(\bullet\) A 20.0 kg rock slides on a rough horizontal surface at 8.00 \(\mathrm{m} / \mathrm{s}\) and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is \(0.200 .\) What average thermal power is produced as the rock stops?

\(\bullet\) \(\bullet\) A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the mass of the insect to be \(10 \mathrm{g},\) and assume the wings move an average downward distance of 1.0 \(\mathrm{cm}\) during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.

\(\bullet\) \(\bullet\) A spring is 17.0 \(\mathrm{cm}\) long when it is lying on a table. One end is then attached to a hook and the other end is pulled by a force that increases to 25.0 \(\mathrm{N}\) , causing the spring to stretch to a length of 19.2 \(\mathrm{cm} .\) (a) What is the force constant of this spring? (b) How much work was required to stretch the spring from 17.0 \(\mathrm{cm}\) to 19.2 \(\mathrm{cm}\) (c) How long will the spring be if the 25 \(\mathrm{N}\) force is replaced by a 50 \(\mathrm{N}\) force?

\(\bullet\) Tall Pacific Coast redwood trees (Sequoia sempervirens) can reach heights of about 100 \(\mathrm{m} .\) If air drag is negligibly small, how fast is a sequoia cone moving when it reaches the ground if it dropped from the top of a 100 \(\mathrm{m}\) tree?

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