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An airplane has a mass of \(3.1 \times 10^{4} \mathrm{~kg}\) and takes off under the influence of a constant net force of \(3.7 \times 10^{4} \mathrm{~N}\). What is the net force that acts on the plane's \(78-\mathrm{kg}\) pilot?

Short Answer

Expert verified
The net force on the pilot is approximately 92.82 N.

Step by step solution

01

Identify Given Information

We are given the mass of the airplane as \(3.1 \times 10^{4} \mathrm{~kg}\) and the net force acting on the airplane as \(3.7 \times 10^{4} \mathrm{~N}\). We also know the mass of the pilot is \(78 \mathrm{~kg}\).
02

Understand the Problem Context

The problem asks for the net force acting on the pilot who is inside the plane. It's important to realize that if the airplane experiences a net force, everything inside it, including the pilot, experiences the same acceleration as the airplane. Consequently, the net force on the pilot depends on the acceleration of the plane and the pilot's mass.
03

Calculate the Acceleration of the Airplane

Use Newton's second law to calculate the acceleration of the airplane. Newton's second law is given by:\[ F = ma \]Where \(F\) is the force, \(m\) is the mass, and \(a\) is the acceleration. Rearrange to find the acceleration:\[ a = \frac{F}{m} = \frac{3.7 \times 10^{4} \mathrm{~N}}{3.1 \times 10^{4} \mathrm{~kg}} \]
04

Compute the Division

Calculate the acceleration:\[ a = \frac{3.7 \times 10^{4}}{3.1 \times 10^{4}} \approx 1.19 \mathrm{~ms^{-2}} \]
05

Calculate the Net Force on the Pilot

Now that we have the acceleration, use it to find the net force on the pilot using the same Newton's second law:\[ F_{pilot} = m_{pilot} \times a \]\[ F_{pilot} = 78 \mathrm{~kg} \times 1.19 \mathrm{~ms^{-2}} \]
06

Compute the Final Force

Calculate the force on the pilot:\[ F_{pilot} = 78 \times 1.19 \approx 92.82 \mathrm{~N} \]

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Key Concepts

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

Net Force
In physics, understanding the concept of "net force" is crucial for analyzing motion. Net force refers to the total sum of all forces acting on an object. Imagine pushing a box on a flat surface. You apply force in one direction, and friction acts in the opposite direction. The combined effect of these forces determines the box's motion.
For the airplane in our exercise, the net force is provided as a constant value, which means it is the vector sum of all forces acting in different directions. When forces are perfectly balanced, the net force is zero, resulting in no change in motion. However, if there's an unbalanced force, like in the case of the airplane, it results in an acceleration according to Newton's Second Law.
  • The net force acting on the airplane is \(3.7 \times 10^{4} \mathrm{~N}\).
  • This net force is responsible for accelerating the airplane and everything inside it.

Knowing how to determine or calculate net force can simplify solving dynamics problems, as it offers insight into how objects move.
Acceleration
Acceleration is the rate of change of velocity over time. It tells us how fast an object speeds up or slows down. Newton's Second Law shows the relationship between an object's mass, its acceleration, and the applied net force.
Mathematically, this is described by the formula \( F = ma \) where \(F\) is force, \(m\) is mass, and \(a\) is acceleration.
For the airplane, calculating the acceleration helps understand how quickly it can increase in speed:
  • Given the force of \( 3.7 \times 10^{4} \text{ N} \) and the airplane's mass of \( 3.1 \times 10^{4} \text{ kg} \), the acceleration is calculated as:

\[ a = \frac{3.7 \times 10^{4} \text{ N}}{3.1 \times 10^{4} \text{ kg}} = 1.19 \text{ m/s}^{2} \]
This acceleration means the airplane and everything inside, like the pilot, will speed up by \( 1.19 \text{ m/s}^2 \) every second. Acceleration plays a huge role in determining forces within systems, especially in scenarios involving motion.
Mass and Force Calculation
The connection between mass and force is at the heart of Newton's Second Law. To calculate the force acting on an object, knowing its mass and acceleration is essential. In simple terms, the net force is equal to the mass times the acceleration: \( F = ma \).
Let's apply this to the airplane's pilot, who has a mass of 78 kg.
  • With the acceleration calculated at \(1.19\text{ m/s}^2\), the net force on the pilot is:
\[ F_{pilot} = 78 \text{ kg} \times 1.19 \text{ m/s}^{2} = 92.82 \text{ N} \]
This force explains how much influence the motion of the airplane has on the pilot inside. It's a straightforward technique to break problems down and successfully apply physics principles. Always remember that mass and force calculations are interconnected, offering insight into the effects of forces on motion.

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Most popular questions from this chapter

The principles used to solve this problem are similar to those in Multiple- Concept Example 17 . A \(205-\mathrm{kg}\) log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at \(30.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is \(0.900,\) and the log has an acceleration of \(0.800 \mathrm{~m} / \mathrm{s}^{2}\). Find the tension in the rope.

A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the earth and the moon. The distance between the earth and the moon is \(3.85 \times 10^{8} \mathrm{~m}\), and the mass of the earth is 81.4 times as great as that of the moon.

A \(325-\mathrm{kg}\) boat is sailing \(15.0^{\circ}\) north of east at a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Thirty seconds later, it is sailing \(35.0^{\circ}\) north of east at a speed of \(4.00 \mathrm{~m} / \mathrm{s}\). During this time, three forces act on the boat: a \(31.0-\mathrm{N}\) force directed \(15.0^{\circ}\) north of east (due to an auxiliary engine), a 23.0-N force directed \(15.0^{\circ}\) south of west (resistance due to the water), and \(\vec{F}_{W}\) (due to the wind). Find the magnitude and direction of the force \(\overrightarrow{\mathrm{F}}_{\mathrm{W}} .\) Express the direction as an angle with respect to due east.

Several people are riding in a hot-air balloon. The combined mass of the people and balloon is \(310 \mathrm{~kg} .\) The balloon is motionless in the air, because the downward-acting weight of the people and balloon is balanced by an upward-acting "buoyant" force. If the buoyant force remains constant, how much mass should be dropped overboard so the balloon acquires an upward acceleration of \(0.15 \mathrm{~m} / \mathrm{s}^{2} ?\)

A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 965 and \(1510 \mathrm{~N}\), respectively. The acceleration of the cable is \(0.620 \mathrm{~m} / \mathrm{s}^{2},\) upward. What is the tension in the cable (a) below the worker and (b) above the worker?

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