/*! 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 111 Astronauts and test pilots exper... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Astronauts and test pilots experience extreme accelerations lasting for very short periods. NASA has been studying the effects of such accelerations. In \(1954,\) one test subject was brought to rest in a water trough from a horizontal velocity of \(286 \mathrm{~m} / \mathrm{s}\) over a distance of \(112.7 \mathrm{~m}\). Assume constant acceleration and a mass of \(77.0 \mathrm{~kg}\). (a) Draw a force diagram for this subject while slowing. (b) What net force was acting during the slow-down? Express your answer in newtons and as a multiple of the subject's weight. (c) For how much time did this force act?

Short Answer

Expert verified
The net force was approximately \(-27,939.45 \mathrm{~N}\), about 37 times the astronaut's weight, and the force acted for approximately 0.788 seconds.

Step by step solution

01

Understand the scenario

The problem involves an astronaut coming to a stop from an initial velocity of \(286 \mathrm{~m/s}\) to a final velocity of \(0 \mathrm{~m/s}\) over a distance of \(112.7 \mathrm{~m}\). We assume constant acceleration.
02

Draw the force diagram

For the force diagram, consider the forces acting on the astronaut: the forward velocity being opposed by the deceleration force. There is no vertical motion, hence the only vertical force is gravity (\(mg\)) balanced by a normal force if considered. The key horizontal force is the deceleration force.
03

Calculate the acceleration

Use the kinematic equation: \( v^2 = u^2 + 2as \), where \(v = 0 \mathrm{~m/s}\), \(u = 286 \mathrm{~m/s}\), and \(s = 112.7 \mathrm{~m}\). Solve for \(a\): \[ 0 = (286)^2 + 2a(112.7) \]\[ a = -\frac{(286)^2}{2 \times 112.7} \approx -362.85 \mathrm{~m/s^2} \]
04

Determine the net force

The net force can be calculated using Newton's second law, \( F = ma \). \( m = 77.0 \mathrm{~kg}\), and \( a = -362.85 \mathrm{~m/s^2} \).\[ F = 77.0 \times (-362.85) = -27939.45 \mathrm{~N} \] (The negative sign indicates the force is in the opposite direction of motion.)
05

Express the net force as a multiple of weight

The astronaut's weight is \( W = mg = 77.0 \times 9.8 = 754.6 \mathrm{~N} \).The net force as a multiple of the weight is: \[ \frac{|-27939.45|}{754.6} \approx 37.02 \]
06

Calculate the duration of the force

Use the equation \( v = u + at \) to find the time \( t \): \[ 0 = 286 + (-362.85)t \]\[ t = \frac{286}{362.85} \approx 0.788 \mathrm{~s} \]

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 91Ó°ÊÓ!

Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the causes of the motion. In this scenario, we are examining how the astronaut comes to a stop over a set distance in a water trough.

The key formula used here is the kinematic equation:
  • \( v^2 = u^2 + 2as \)
where:
  • \( v \) is the final velocity
  • \( u \) is the initial velocity
  • \( a \) is the acceleration
  • \( s \) is the distance
To solve for acceleration \( a \), we rearrange the formula:
  • \( a = \frac{v^2 - u^2}{2s} \)

Kinematics involves understanding how variables like velocity, distance, and time interact, making it a fundamental concept to solve problems involving motion.
Newton's Laws
Newton's Laws of Motion are key principles in understanding forces and motion. In this exercise, Newton’s Second Law is pivotal. It states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
  • \( F = ma \)
This allows us to calculate the force acting on the astronaut as he decelerates in the water trough.
Newton's First Law also supports this scenario by demonstrating that an object in motion, such as the astronaut, will continue moving at a constant speed unless acted upon by an external force, which in this situation is the deceleration force.
Newton’s Laws provide a framework to predict how objects will move or stop moving under various forces, crucial for solving real-world physics problems.
Force Diagrams
A force diagram, or free body diagram, is a visual tool that helps analyze the forces acting on an object. In this scenario, the diagram must show the forces on the astronaut as he comes to a stop.

Key components include:
  • Horizontal forces showing deceleration, opposing the motion
  • Vertical forces, such as gravity pulling him down, balanced by a normal force from the water
Drawing these forces helps in understanding how the forces interact and the resultant vector, which in this case leads to the astronaut slowing down.
Force diagrams simplify complex situations, breaking them down into manageable parts to see the overall effect of different forces.
Acceleration Calculations
Accelerations describe the rate at which an object changes its velocity. For constant acceleration, the rate remains the same over the time period considered.

Using the kinematic equation:
  • \( v = u + at \),
you can solve for time \( t \) when acceleration \( a \), initial velocity \( u \), and final velocity \( v \) are known.

Ilustrating with the astronaut's scenario,
  • \( t = \frac{v - u}{a} \)
This calculation shows how quickly an object can come to a full stop under constant deceleration, providing insight into the dynamics of motion.
Mastering acceleration calculations helps predict and understand motion in varied situations, like the rapid deceleration experienced by the astronaut.

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

A truck carries a \(3.0-\mathrm{kg}\) box of Washington apples on its flat bed. The coefficient of static friction between box and truck bed is 0.38. (a) What is the maximum acceleration for the truck on a level road if the box isn't to slide? (b) Repeat part (a) assuming the truck is driving toward Snoqualmie Pass on a \(4.5^{\circ}\) upward incline.

A 365 -kg bobsled starts down an icy slope inclined at \(3.4^{\circ}\). (a) If the sled's acceleration along the incline is \(0.51 \mathrm{~m} / \mathrm{s}^{2},\) what is the coefficient of kinetic friction? (b) A velocity- dependent air drag force from the air is directed opposite the sled's velocity. Eventually, the drag reduces the acceleration to zero. What's the magnitude of the drag force at that time?

A 0.17 -kg cue ball rests on the pool table. It's struck by a cue stick applying force \(\vec{F}_{1}=15 \mathrm{~N} \hat{\imath}+36 \mathrm{~N} \hat{\jmath}\). (a) Determine the magnitude and direction of the ball's acceleration. (b) If the force was applied for \(0.015 \mathrm{~s}\), what's the ball's final speed?

A child ties a rock to a string and whirls it around in a horizontal circle. (a) The string cannot be perfectly horizontal. Why not? Explain using a force diagram. (b) Assuming a \(1.22-\mathrm{m}\) -long string making \(25^{\circ}\) angle below the horizontal, find (c) the speed of the rock and (d) the period of its uniform circular motion.

An elevator undergoes constant upward acceleration, taking it from rest to \(13.4 \mathrm{~m} / \mathrm{s}\) in \(4.0 \mathrm{~s}\). An \(80-\mathrm{kg}\) man stands on a scale in the elevator. (a) Make a force diagram for the man. (b) What does the scale read during the upward acceleration? (c) Nearing the top floor, the elevator slows from \(13.4 \mathrm{~m} / \mathrm{s}\) to rest in the last \(22.4 \mathrm{~m}\) of upward travel. Assuming constant acceleration, what's the scale reading during this period?

See all solutions

Recommended explanations on Physics Textbooks

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.