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Chapter 5: Problem 13

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/s虏 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 \( t \):\[ 0 = 86.39 + (-4608.2) \times t \]\[ t = \frac{86.39}{4608.2} \approx 0.0187 \text{ s} \]

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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.

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Most popular questions from this chapter

A 3.00-kg box that is several hundred meters above the earth's surface is suspended from the end of a short vertical rope of negligible mass. A time- dependent upward force is applied to the upper end of the rope and results in a tension in the rope of \(T(t) =\) (36.0 N/s)\(t\). The box is at rest at \(t =\) 0. The only forces on the box are the tension in the rope and gravity. (a) What is the velocity of the box at (i) \(t =\) 1.00 s and (ii) \(t =\) 3.00 s? (b) What is the maximum distance that the box descends below its initial position? (c) At what value of \(t\) does the box return to its initial position?

Block \(A\), with weight \(3w\), slides down an inclined plane \(S\) of slope angle 36.9\(^{\circ}\) at a constant speed while plank \(B\), with weight \(w\), rests on top of A. The plank is attached by a cord to the wall \(\textbf{(Fig. P5.97).}\) (a) Draw a diagram of all the forces acting on block \(A\). (b) If the coefficient of kinetic friction is the same between \(A\) and \(B\) and between \(S\) and \(A\), determine its value.

A 2540-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by \(v(t) = At + Bt^2\), where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50 m/s\(^2\) at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m>s. (a) Determine \(A\) and \(B\), including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

An 8.00-kg box sits on a ramp that is inclined at 33.0\(^\circ\) above the horizontal. The coefficient of kinetic friction between the box and the surface of the ramp is \(\mu_k =\) 0.300. A constant \(horizontal\) force \(F =\) 26.0 N is applied to the box (\(\textbf{Fig. P5.73}\)), and the box moves down the ramp. If the box is initially at rest, what is its speed 2.00 s after the force is applied?

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?
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