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Genesis Crash. On September \(8,2004,\) the Genesis spacecraft crashed in the Utah desert because its parachute did not open. The \(210 \mathrm{~kg}\) capsule hit the ground at \(311 \mathrm{~km} / \mathrm{h}\) and penetrated the soil to a depth of \(81.0 \mathrm{~cm}\). (a) What was its acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) 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
The answers can be found by doing the arithmetic calculations.

Step by step solution

01

Convert the given units

First, convert the given units into compatible ones. Convert kilometers/hours into meters/seconds for velocity (1km = 1000m, 1 hour = 3600 seconds), and centimeters into meters for depth (1cm = 0.01m). \[311 \ km/h = 311000\ m/3600\ s = 86.39 \ m/s\] \[81\ cm = 81*0.01\ m = 0.81\ m \]
02

Find acceleration

We will use the kinematic equation (assuming acceleration (a) is constant)\[v_f^2 = v_i^2 + 2as\] Here, the final velocity (\(v_f\)) after the spacecraft has crashed is 0 (it's at rest), the initial velocity (\(v_i\)) is 86.39 m/s (which we found in Step 1), the distance (s) is 0.81 m (depth of penetration which we also found in Step 1). By substituting these values, we can find acceleration.\[0 = (86.39)^2 + 2*a*0.81\] Now solve for 'a' - this will give the magnitude of deceleration (it's deceleration as the spacecraft is slowing down).
03

Find force

To find the force that the ground exerted on the spacecraft during the crash, we shall use Newton's second law of motion: 'force = mass * acceleration'. The mass of the capsule is given to be 210 kg and we've found acceleration in the previous step.
04

How long did the force last?

To find out how long the decelerating force acted on the spacecraft, we use one of the kinematic equations: \(t = (v_f - v_i) / a\). We already know the values of \(v_f\), \(v_i\) and \(a\) from previous steps, so by substiting these values, we can find 't' (time).
05

Calculation

Now, substitute the values into the equations derived in the earlier steps to get numerical answers.

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Key Concepts

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

Constant Acceleration
In physics, constant acceleration refers to a situation where an object's velocity changes at a steady rate over time. This is a key concept when analyzing motion because it greatly simplifies calculations and predictions. For example, in the case of the Genesis spacecraft, the problem assumes that the craft experienced constant acceleration, or in other terms, unchanging deceleration, during its unfortunate descent.

This is essential for utilizing the kinematic equations which are designed to work under the assumption of constant acceleration. Under these conditions, acceleration can be calculated using the known values of initial velocity, final velocity (which is zero if the spacecraft has come to a stop), and the distance over which the change in velocity occurred.

Constant acceleration is often denoted by the symbol 'a' in equations and can be positive when the object is speeding up or negative in cases of deceleration when a force is applied in the opposite direction of motion.
Newton's Second Law of Motion
Newton's second law of motion is a fundamental principle in physics that states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). This law explains the relationship between an object's motion and the forces acting upon it.

When applied to the Genesis spacecraft problem, this law allows us to understand the force the ground exerted on the capsule during impact. Since we know the capsule's mass and can compute its constant acceleration during the crash, we can calculate the force. It's vital to remember that this force is a vector quantity, which means it has both magnitude and direction, providing deeper insight into how the spacecraft came to a halt.
Kinematic Equations
Kinematic equations are a set of four equations that describe the motion of objects under the influence of constant acceleration. These equations relate the displacement, initial and final velocities, acceleration, and time of an object's motion.

For the Genesis spacecraft, the kinematic equation \(v_f^2 = v_i^2 + 2as\) was used to compute its acceleration during the crash. This particular equation is useful for solving problems where time isn't directly involved.

However, to determine the crash's duration, another kinematic equation \(t = (v_f - v_i) / a\) helps us calculate the time over which the deceleration took place. These equations are powerful tools for problem-solving in kinematics, helping to break down complex motion into understandable parts.
Deceleration
Deceleration is the term used to describe negative acceleration, meaning it is the reduction in the velocity of an object over time. It is acceleration in the direction opposite to the direction of the velocity. Deceleration occurs when forces act contrary to the direction of motion, causing an object to slow down.

In the context of the Genesis spacecraft, deceleration is what occurred as the spacecraft hit the ground. It went from a certain speed to a complete stop over a particular distance. This deceleration can be calculated using kinematic equations as long as it remains constant. Not only is this concept crucial for understanding crashes like that of the spacecraft, but it is also applicable in everyday situations, such as a car coming to a stop when the brakes are applied.

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Most popular questions from this chapter

A \(2.00 \mathrm{~kg}\) box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope, and the box moves upward with a velocity magnitude that varies in time according to \(v(t)=\left(2.00 \mathrm{~m} / \mathrm{s}^{2}\right) t+\left(0.600 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\). What is the tension in the rope when the velocity of the box is \(9.00 \mathrm{~m} / \mathrm{s} ?\)

Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate \(A\) has mass \(m_{A}\), and crate \(B\) has mass \(m_{B}\). The coefficient of kinetic friction between each crate and the surface is \(\mu_{\mathrm{k}} .\) The crates are pulled to the right at constant velocity by a horizontal force \(\overrightarrow{\boldsymbol{F}}\). Draw one or more free-body diagrams to calculate the following in terms of \(m_{A}, m_{B},\) and \(\mu_{\mathrm{k}}:\) (a) the magnitude of \(\overrightarrow{\boldsymbol{F}}\) and \((\mathrm{b})\) the tension in the rope connecting the blocks.

A block with mass \(m_{1}\) is placed on an inclined plane with slope angle \(\alpha\) and is connected to a hanging block with mass \(m_{2}\) by a cord passing over a small, friction less pulley (Fig. P5.74). The coefficient of static friction is \(\mu_{\mathrm{s}}\), and the coefficient of kinetic friction is \(\mu_{\mathrm{k}}\). (a) Find the value of \(m_{2}\) for which the block of mass \(m_{1}\) moves up the plane at constant speed once it is set in motion. (b) Find the value of \(m_{2}\) for which the block of mass \(m_{1}\) moves down the plane at constant speed once it is set in motion. (c) For what range of values of \(m_{2}\) will the blocks remain at rest if they are released from rest?

A box of bananas weighing \(40.0 \mathrm{~N}\) rests on a horizontal surface. The coefficient of static friction between the box and the surface is \(0.40,\) and the coefficient of kinetic friction is \(0.20 .\) (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of \(6.0 \mathrm{~N}\) to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of \(18.0 \mathrm{~N},\) what is the magnitude of the friction force and what is the box's acceleration?

The cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of \(100 \mathrm{~m}\). Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every \(60.0 \mathrm{~s}\) ). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs \(882 \mathrm{~N}\) at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

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