/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 BIO (a) A 75-kg man steps out a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 69

BIO (a) A 75-kg man steps out a window and falls (from rest) \(1.0 \mathrm{~m}\) to a sidewalk. What is his speed just before his feet strike the pavement? (b) If the man falls with his knees and ankles locked, the only cushion for his fall is an approximately \(0.50-\mathrm{cm}\) give in the pads of his feet. Calculate the average force exerted on him by the ground during this \(0.50 \mathrm{~cm}\) of travel. This average force is sufficient to cause damage to cartilage in the joints or to break bones.

Short Answer

Expert verified
The speed at which the man hits the ground is found to be 4.43 \(m/s\). The average force exerted on him by the ground is around 22.3 kN, which is dangerous and can cause physical harm.

Step by step solution

01

Determine the man's speed when he hits the ground

Using the kinematic equation \(v^2 = u^2 + 2gs\) where \(v\) is final velocity, \(u\) is initial velocity(0 in this case as the man drops from rest), \(g\) is acceleration due to gravity(9.8 \(m/s^2\)) and \(s\) is distance travelled (1.0 m), calculate the final velocity \(v\) of the man as he hits the ground. This equation gives \(v = \sqrt{u^2 + 2gs}\).
02

Calculate the man's deceleration

Find out how fast the man decelerated when he hits the ground. Use the same kinematic equation with initial velocity \(u\) as the man's speed as he hits the ground (that was calculated in Step 1), final velocity \(v\) as 0 (because he stops when he hits the ground) and distance \(s\) as 0.50 cm (the give in the pads of his feet when he hits the ground), converted into meters (= 0.50/100 m). Solve this equation for acceleration \(a\), and the result will be a negative number, indicating deceleration.
03

Calculate the average force exerted on the man by the ground

Use Newton's Second Law, which states that force is equal to mass times acceleration (\(F = ma\)). Apply the man's mass (75 kg) and the negative acceleration found in Step 2 to solve this equation for force \(F\). This will represent the force that the ground exerts on the man, acting upwards.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Free Fall Motion
When an object falls freely under the influence of gravity alone, it experiences free fall motion. In this scenario, no other forces, like air resistance, are acting against it. The motion is determined entirely by gravity. For the man in the exercise, he starts from rest, meaning his initial speed is zero.

We use the kinematic equation, which is helpful in scenarios like this:
  • \(v^2 = u^2 + 2gs\)
This formula allows us to find the final velocity \(v\) just before the man hits the ground. Here, \(u = 0\) (initial velocity), \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity), and \(s = 1.0 \, \text{m}\) (distance fallen).

Solving this, we calculate the speed at the point of impact.
Acceleration Due to Gravity
Gravity is a force that pulls objects toward the center of Earth. This force results in acceleration, known as acceleration due to gravity, denoted by \(g\).

  • The standard value of \(g\) is approximately \(9.8 \, \text{m/s}^2\).
  • It does not depend on the mass of the falling object.
In the exercise, the constant acceleration due to gravity helps us compute how fast the man's speed increases as he falls. Since no other forces are acting (like air resistance), the acceleration remains constant throughout the fall.
Newton's Second Law
Newton's Second Law connects force, mass, and acceleration in the equation \(F = ma\).

After calculating how fast the man decelerates when hitting the ground, we can find the force exerted by applying his mass and this deceleration.

  • The man's mass is \(75 \, \text{kg}\).
  • The acceleration (or deceleration in this context) is calculated from the velocity change over the distance of \(0.50 \, \text{cm}\) as he comes to a stop.
By substituting these values into the formula, we determine the average force the ground exerts upward on the man. This force is significant enough to cause potential injuries.
Impact Force Calculation
When the man hits the ground, he comes to a stop quickly, compressing the padding of his feet by \(0.50 \, \text{cm}\). Calculating impact force involves understanding how this rapid deceleration translates into force.

  • First, we convert \(0.50 \, \text{cm}\) to meters \((0.50/100)\) to work within the metric system continuously.
  • We use the man's initial impact speed calculated earlier as the initial velocity for this short stop and find the deceleration needed for him to stop in this distance.
  • Applying the resulting negative acceleration into Newton鈥檚 Second Law \(F = ma\), we calculate the force exerted by the ground.
This force calculation is critical to understanding the potential harms of such a fall.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

BIO The chin-up is one exercise that can be used to strengthen the biceps muscle. This muscle can exert a force of approximately \(800 \mathrm{~N}\) as it contracts a distance of \(7.5 \mathrm{~cm}\) in a \(75-\mathrm{kg}\) male. \({ }^{3}\) How much work can the biceps muscles (one in cach arm) perform in a single contraction? Compare this amount of work with the energy required to lift a \(75-\mathrm{kg}\) person \(40 \mathrm{~cm}\) in performing a chin-up. Do you think the biceps muscle is the only muscle involved in performing a chin- up?

A truck travels uphill with constant velocity on a highway with a \(7.0^{\circ}\) slope. A \(50-\mathrm{kg}\) package sits on the floor of the back of the truck and does not slide, due to a static frictional force. During an interval in which the truck travels \(340 \mathrm{~m}\), (a) what is the net work done on the package? What is the work done on the package by (b) the force of gravity, (c) the normal force, and (d) the friction force?

\(\mathrm{S}\) A projectile of mass \(m\) is fired horizontally with an initial speed of \(v_{0}\) from a height of \(h\) above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of \(m, v_{0}, h\), and \(g\) : (a) the work done by the force of gravity on the projectile, (b) the change in kinetic energy of the projectile since it was fired, and (c) the final kinetic energy of the projectile. (d) Are any of the answers changed if the initial angle is changed?

BIO Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is 1 kilocalorie, which we define in Chapter 11 as \(1 \mathrm{kcal}=4186 \mathrm{~J}\). Metabolizing 1 gram of fat can release \(9.00 \mathrm{kcal}\). A student decides to try to lose weight by exercising. She plans to run up and down the stairs in a football stadium as fast as she can and as many times as neccssary. Is this in itself a practical way to lose weight? To evaluate the program, suppose she runs up a flight of 80 steps, each \(0.150 \mathrm{~m}\) high, in \(65.0 \mathrm{~s}\). For simplicity, ignore the energy she uses in coming down (which is small). Assume that a typical efficiency for human muscles is \(20.0 \%\). This means that when your body converts \(100 \mathrm{~J}\) from metabolizing fat, \(20 \mathrm{~J}\) goes into doing mechanical work (here, climbing stairs). The remainder goes into internal energy. Assume the student's mass is \(50.0 \mathrm{~kg}\). (a) How many times must she run the flight of stairs to lose 1 pound of fat? (b) What is her average power output, in watts and in horsepower, as she is running up the stairs?

Q C Starting from rest, a \(5.00-\mathrm{kg}\) block slides \(2.50 \mathrm{~m}\) down a rough \(30.0^{\circ}\) incline. The coefficient of kinetic friction between the block and the incline is \(\mu_{k}=0.436\). Determine (a) the work done by the force of gravity, (b) the work done by the friction force between block and incline, and (c) the work done by the normal force. (d) Qualitatively, how would the answers change if a shorter ramp at a steeper angle were used to span the same vertical height?
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Nuclear Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Force

Read Explanation

Wave Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.