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A spaceship lifts off vertically from the Moon, where \(g=1.6 \mathrm{~m} / \mathrm{s}^{2} .\) If the ship has an upward acceleration of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) as it lifts off, what is the magnitude of the force exerted by the ship on its pilot, who weighs \(735 \mathrm{~N}\) on Earth?

Short Answer

Expert verified
The force exerted by the ship on the pilot is 195 N.

Step by step solution

01

Calculate the Pilot's Mass

On Earth, the weight of the pilot is given by the formula \( W = mg \), where \( g = 9.8 \mathrm{~m/s^2} \). The pilot's weight on Earth is \( W = 735 \mathrm{~N} \). We can find the mass \( m \) by rearranging the formula: \( m = \frac{W}{g} = \frac{735}{9.8} \approx 75 \mathrm{~kg} \).
02

Determine the Net Acceleration on the Moon

The net acceleration of the spaceship on the Moon is the sum of the gravitational acceleration and the spaceship's acceleration. Since the gravitational acceleration on the Moon is \( 1.6 \mathrm{~m/s^2} \) and the upward acceleration is \( 1.0 \mathrm{~m/s^2} \), the net acceleration \( a_{net} = g_{moon} + a = 1.6 + 1.0 = 2.6 \mathrm{~m/s^2} \).
03

Calculate the Force Exerted on the Pilot

Using Newton's second law, \( F = ma \), where \( m \) is the pilot's mass and \( a \) is the net acceleration on the Moon, we find the force exerted on the pilot: \( F = 75 \times 2.6 = 195 \mathrm{~N} \).

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Key Concepts

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

Newton's second law
Newton's second law is a fundamental principle in physics that describes the relationship between an object's mass, acceleration, and the force applied to it. The law is often expressed with the equation \( F = ma \), where \( F \) is the force applied to the object, \( m \) is the mass of the object, and \( a \) is the acceleration it experiences as a result of the force. This equation helps us understand how the force acting on an object is proportional to both its mass and the acceleration it undergoes.

For example, if you increase the force applied to an object, the acceleration increases assuming the object's mass remains constant. Conversely, if the mass increases while the force remains the same, the acceleration decreases. This concept is crucial in solving problems involving moving objects, such as the spaceship in the exercise.
  • Mass \(m\) is the measure of the amount of matter in the object.
  • Acceleration \(a\) refers to the change in velocity per unit time.
  • Newton's second law explains how varying force alters acceleration depending on the object's mass.
Gravitational acceleration
Gravitational acceleration is the acceleration that an object experiences due to the gravitational force exerted by a celestial body, like Earth or the Moon. On Earth, the gravitational acceleration \( g \) is approximately \( 9.8 \, \mathrm{m/s^2} \). However, this value is different on other celestial bodies because gravitational force depends on both mass and the distance to the body's center of mass.

In the problem, the spaceship is on the Moon, where gravitational acceleration is much weaker at \( 1.6 \, \mathrm{m/s^2} \). This difference affects how objects move when subjected to gravitational forces on the Moon. The pilot's weight experienced due to gravitational force is notably less than what they experience on Earth.
  • On Earth: \( g = 9.8 \, \mathrm{m/s^2} \)
  • On the Moon: \( g_{moon} = 1.6 \, \mathrm{m/s^2} \)
  • Weight is calculated as \( W = mg \), illustrating the direct effect of gravitational acceleration on perceived weight.
Net acceleration
Net acceleration describes the total acceleration experienced by an object when multiple forces are acting on it simultaneously. It accounts for both the external forces and any opposing forces, such as gravity. To find net acceleration, we add the upward acceleration provided by the spaceship and the gravitational acceleration acting downward.

In our scenario, the spaceship's engines provide an upward acceleration of \( 1.0 \, \mathrm{m/s^2} \). When combined with the Moon's gravitational pull \( 1.6 \, \mathrm{m/s^2} \), we calculate the net acceleration as \( 2.6 \, \mathrm{m/s^2} \).
  • Net acceleration \( a_{net} \) is a resultant of different acceleration vectors acting on the object.
  • Calculated as the sum of forces in specific directions, considering positive for up and negative for down.
  • Informs the total increase in velocity perceived by the object, crucial for solving dynamics problems.

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Most popular questions from this chapter

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