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Chapter 5: Problem 38

\(\cdot\) Suppose a rocket launches with an acceleration of \(30.5 \mathrm{m} / \mathrm{s}^{2}\) What is the apparent weight of an \(92-\mathrm{kg}\) astronaut aboard this rocket?

Short Answer

Expert verified
Apparent weight is 3812 N.

Step by step solution

01

Understanding the Problem

We need to find the apparent weight of an astronaut of mass 92 kg who is in a rocket accelerating at 30.5 m/s虏. Apparent weight is the normal force experienced by the astronaut due to the rocket's acceleration.
02

Calculating True Weight

The true weight of an astronaut is calculated using the formula: \[ W = mg \]where \( m = 92 \text{ kg} \) is the mass and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Putting these values into the formula gives:\[ W = 92 \times 9.8 = 901.6 \text{ N} \].

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Key Concepts

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

Rocket Acceleration
When we talk about rocket acceleration, we are describing how quickly a rocket increases its velocity. This change in speed is influenced by the forces acting on the rocket. In our example, the rocket is accelerating at a rate of 30.5 meters per second squared (m/s虏). To put this number in perspective, this is similar to a car reaching a high speed in a very short time.

Rocket acceleration depends on several factors:
  • Thrust: The force pushing the rocket upwards, overcoming gravity.
  • Mass: The total mass of the rocket, including the fuel and the astronaut.
  • Gravity: The constant pull of Earth trying to pull the rocket back down.
The faster the rocket accelerates, the harder it is pushed against gravity. As the rocket speeds up, passengers inside feel an increased force pressing them towards the floor or seat, known commonly as 'apparent weight'. This is because the additional acceleration adds to the force of gravity.
Normal Force
The concept of normal force is central to understanding weight readings in new environments, like in an accelerating rocket. Normal force (often symbolized as N) is a force exerted perpendicular to the surface of contact. In simple terms, it is the force that counteracts gravity to keep an object stationary relative to a surface.

In our exercise, the normal force is what the astronaut perceives as their weight inside the accelerating rocket. However, due to the rocket's acceleration, the normal force experienced is greater than the astronaut's true weight. Apparent weight is this felt weight, and it's calculated with:\[ N = m(g + a) \]where:
  • \( m \) is the mass of the astronaut (92 kg in our example).
  • \( g \) is the acceleration due to gravity (9.8 m/s虏).
  • \( a \) is the acceleration of the rocket (30.5 m/s虏).
This equation tells us that as the rocket accelerates, the astronaut feels even heavier, as if gravity were stronger.
Mass of Astronaut
The mass of an astronaut plays a vital role in determining their apparent weight in an accelerating rocket. Mass is a measure of how much matter an object contains, and it is measured in kilograms (kg).

For our scenario, the astronaut has a mass of 92 kg. This constant value is used to calculate both the true weight and the apparent weight:
  • True weight is \( 92 \times 9.8 \), which equals 901.6 N.
  • Apparent weight uses the rocket's acceleration in the formula \( N = m(g + a) \).
Mass is important because it directly influences the astronaut's weight. While the actual mass remains unchanged regardless of location, how heavy the astronaut feels depends on the surrounding forces, including those from acceleration. This change in apparent weight is what astronauts experience during rocket launches and other high-acceleration scenarios.

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Most popular questions from this chapter

\- CE Predict/Explain A small car collides with a large truck. (a) Is the magnitude of the force experienced by the car greater than, less than, or equal to the magnitude of the force experienced by the truck? (b) Choose the best explanation from among the following: I. Action-reaction forces always have equal magnitude. II. The truck has more mass, and hence the force exerted on it is greater. III. The massive truck exerts a greater force on the lightweight car.

Before practicing his routine on the rings, a \(67-\mathrm{kg}\) gymnast stands motionless, with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of \(24^{\circ}\) above the horizontal. (a) If the force exerted by the rings on each arm has a magnitude of \(290 \mathrm{N},\) and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet? (b) If the angle his arms make with the horizontal is greater that \(24^{\circ},\) and everything else remains the same, is the force exerted by the floor on his feet greater than, less than, or the same as the value found in part (a)? Explain.

An air-track cart of mass \(m_{1}=0.14 \mathrm{kg}\) is moving with a speed \(v_{0}=1.3 \mathrm{m} / \mathrm{s}\) to the right when it collides with a cart of mass \(\mathrm{m}_{2}=0.25 \mathrm{kg}\) that is at rest. Each cart has a wad of putty on its bumper, and hence they stick together as a result of their collision. Suppose the average contact force between the carts is \(F=1.5 \mathrm{N}\) during the collision. (a) What is the acceleration of cart \(1 ?\) Give direction and magnitude. (b) What is the acceleration of cart \(2 ?\) Give direction and magnitude. (c) How long does it take for both carts to have the same speed? (Once the carts have the same speed the collision is over and the contact force vanishes.) (d) What is the final speed of the carts, \(v_{\mathrm{f}} ?\) (e) Show that \(m_{1}{ }^{2} v_{0}\) is equal to \(\left(m_{1}+m_{2}\right) v_{1}\). (We shall investigate the significance of this result in Chapter 9.3

A driver who does not wear a seatbelt continues to move forward with a speed of \(18.0 \mathrm{m} / \mathrm{s}\) (due to inertia) until something solid like the steering wheel is encountered. The driver now comes to rest in a much shorter distance - perhaps only a few centimeters. Find the magnitude of the net force acting on a \(65.0-\mathrm{kg}\) driver who is decelerated from \(18.0 \mathrm{m} / \mathrm{s}\) to rest in \(5.00 \mathrm{cm} .\) A. \(3240 \mathrm{N}\) B. \(1.17 \times 10^{4} \mathrm{N}\) C. \(2.11 \times 10^{5} \mathrm{N}\) D. \(4.21 \times 10^{5} \mathrm{N}\)

A 92 -kg water skier floating in a lake is pulled from rest to a speed of \(12 \mathrm{m} / \mathrm{s}\) in a distance of \(25 \mathrm{m}\). What is the net force exerted on the skier, assuming his acceleration is constant?
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