Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

On September 8, 2004, the \(Genesis\) spacecraft crashed in the Utah desert because its parachute did not open. The 210-kg capsule hit the ground at 311 km/h and penetrated the soil to a depth of 81.0 cm. (a) What was its acceleration (in m/s\(^2\) and in g's), assumed to be constant, during the crash? (b) What force did the ground exert on the capsule during the crash? Express the force in newtons and as a multiple of the capsule's weight. (c) How long did this force last?

Short Answer

Expert verified
(a) -4608.2 m/sĀ² or -469.6 g's; (b) 968,682 N or 470 times the weight; (c) 0.0187 s.

Step by step solution

01

Convert Units

Before we start, convert the velocity from km/h to m/s. We have:\[\text{Initial Velocity, } v_i = 311 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 86.39 \text{ m/s} \]Also, convert the depth from cm to m:\[\text{Depth, } d = 81.0 \text{ cm} = 0.81 \text{ m} \]
02

Apply the Kinematic Equation

We use the kinematic equation: \[ v_f^2 = v_i^2 + 2ad \]where the final velocity \( v_f = 0 \text{ m/s} \) (as the capsule comes to rest), the initial velocity \( v_i = 86.39 \text{ m/s} \), and the depth \( d = 0.81 \text{ m} \).Rearrange for acceleration \( a \):\[ 0 = (86.39)^2 + 2 \times a \times 0.81 \]\[ a = -\frac{(86.39)^2}{2 \times 0.81} \approx -4608.2 \text{ m/s}^2 \]
03

Convert Acceleration to g's

Express the acceleration in terms of gravity (\( g = 9.81 \text{ m/s}^2 \)):\[ a_{\text{in g's}} = \frac{-4608.2}{9.81} \approx -469.6 \text{ g's} \]
04

Calculate the Force Exerted by the Ground

Using Newton's Second Law, \( F = ma \), find the force:\[ F = 210 \text{ kg} \times 4608.2 \text{ m/s}^2 \approx 968,682 \text{ N} \]
05

Express Force as a Multiple of Weight

The weight \( w \) of the capsule is:\[ w = mg = 210 \times 9.81 = 2060.1 \text{ N} \]The force as a multiple of the capsule's weight is:\[ \frac{968,682 \text{ N}}{2060.1 \text{ N}} \approx 470 \]
06

Find the Duration of the Force

Use the formula \( v_f = v_i + at \) to find the time \( t \):\[ 0 = 86.39 + (-4608.2) \times t \]\[ t = \frac{86.39}{4608.2} \approx 0.0187 \text{ s} \]

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 Vaia!

Key Concepts

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

Acceleration
Acceleration is a key factor in understanding how quickly an object changes its velocity. During a crash scenario, like that of the Genesis spacecraft, the acceleration experienced can be extreme.
This is because the object transitions from a high initial speed to rest over a very short distance.
When everything else is constant, a greater change in velocity over a smaller distance, or shorter time imparts a higher acceleration to the object involved.
To calculate acceleration in the scenario of the spacecraft, we first had to convert the initial velocity to meters per second (m/s).
Using kinematic equations, we input this converted velocity along with the depth to which the spacecraft penetrated the ground.
The primary kinematic equation used was:
  • \[ v_f^2 = v_i^2 + 2ad \]
where:
  • \( v_f \): final velocity (0 m/s)
  • \( v_i \): initial velocity (86.39 m/s)
  • \( a \): acceleration (unknown)
  • \( d \): distance (0.81 m)
Solving for acceleration gives us an extremely high acceleration in meters per second squared (m/s\(^2\)), which indicates a rapid decrease in speed over the crash impact.
This is further expressed in terms of gravitational forces (g's), emphasizing just how extreme this acceleration was.
"g" corresponds to the acceleration due to gravity, approximately 9.81 m/s\(^2\). To convert the calculated acceleration into g's, divide it by 9.81.
Force Calculation
The concept of force is interlinked with that of acceleration and mass in mechanical scenarios, especially in collisions.
It's essentially the "push" or "pull" felt by objects that are interacting.
For the Genesis spacecraft crash, calculating the force exerted by the ground involves understanding Newton's Second Law of Motion and applying it.
We use the established formula:
  • \[ F = ma \]
where:
  • \( F \): force
  • \( m \): mass of the spacecraft (210 kg)
  • \( a \): acceleration (calculated as 4608.2 m/s\(^2\))
By multiplying the mass by the calculated acceleration, we find the overall force exerted during the impact.
The result in newtons (N) is a reflection of just how impactful the crash was.
To contextualize this immense force, we also express it as a multiple of the spacecraft's own weight.
Weight here is the force due to gravity on the spacecraft.
By dividing the force exerted by the ground by the object's weight, we understand just how much greater the crash force was compared to the routine force of gravity acting on the object.
Newton's Second Law
Newton's Second Law of Motion forms the backbone of many calculations in physics, particularly those involving motion and forces.
It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration.
This is mathematically expressed as:
  • \[ F = ma \]
The simplicity of this equation belies its immense utility in diagnosing and understanding motion scenarios.
For the Genesis spacecraft, this law was instrumental in calculating the force acting during the crash.
By linking force, mass, and acceleration, Newton's Second Law allows us to understand the influences an object encounters when it changes its speed or direction.
It provides insight into how external influences can affect a body in motion.
In the case of the spacecraft crash, the high force calculated via this law underscores the dramatic deceleration and severity of the impact experienced.
Understanding and applying Newton's Second Law give us a clearer view of not just magnitudes of forces, but how they shape movement and impact in our physical world.

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

You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 kg. The elevator starts from rest and travels upward with a speed that varies with time according to \(v(t) =\) (3.0 m/s\(^2\))\(t\) \(+\) (0.20 m/s\(^3\))\(t^2\). When \(t =\) 4.0 s, what is the reading on the bathroom scale?

When jumping straight up from a crouched position, an average person can reach a maximum height of about 60 cm. During the jump, the person's body from the knees up typically rises a distance of around 50 cm. To keep the calculations simple and yet get a reasonable result, assume that the \(entire\) \(body\) rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of 60 cm? (b) Draw a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(w\), what force does the ground exert on him or her during the jump?

A curve with a 120-m radius on a level road is banked at the correct angle for a speed of 20 m/s. If an automobile rounds this curve at 30 m/s, what is the minimum coefficient of static friction needed between tires and road to prevent skidding?

A 2.00-kg box is moving to the right with speed 9.00 m/s on a horizontal, frictionless surface. At \(t =\) 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude \(F(t) = (6.00 N/s^2)t^2\). (a) What distance does the box move from its position at \(t =\) 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at \(t =\) 3.00 s?

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates "artificial gravity" at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the "artificial gravity" acceleration to be 9.80 m/s\(^2\)? (b) If the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravity on the Martian surface (3.70 m/s\(^2\)). How many revolutions per minute are needed in this case?

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.

Sign-up for free