/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 Compared with the force her neck... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Compared with the force her neck exerts on her head during the landing, the force her head exerts on her neck is (a) the same; (b) greater; (c) smaller; (d) greater during the first half of the landing and smaller during the second half of the landing.

Short Answer

Expert verified
The force her head exerts on her neck is the same as the force her neck exerts on her head (Option a).

Step by step solution

01

Understand Newton's Third Law

To solve this exercise, the first step is understanding Newton's third law of motion. It states that 'for every action, there is an equal and opposite reaction.' This means that any force exerted on a body will elicit a response from that body of equal magnitude but in the opposite direction.
02

Apply Newton's Third Law to the Stated Problem

Applying this principle to the problem stated, when the girl lands, her head exerts a certain force on her neck. This is the action. The neck, in response, will exert an equal force in the opposite direction. This is the reaction.
03

Conclusion

Given that Newton's third law of motion dictates an equal and opposite reaction for all actions, the force the girl's head exerts on her neck during the landing will be equal in magnitude to the force her neck exerts on her head, albeit in opposite directions.

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Key Concepts

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

Action-Reaction Pairs
In every interaction, forces come in pairs, which are known as action-reaction pairs. Whenever you push against something, it pushes back with equal strength but in the opposite direction. This is a fundamental part of Newton's Third Law, where these pairs of forces are equal in magnitude and opposite in direction.

In the context of our problem, when the girl's head pushes down on her neck as she lands, her neck pushes back up on her head with an equal force. This interaction forms the action-reaction pair:
  • Action: Head exerting force downward on the neck
  • Reaction: Neck exerting force upward on the head
Because these forces are equal and opposite, neither force takes precedence over the other. Understanding this concept helps explain how balanced forces operate in daily activities, such as jumping or even pressing keys on a keyboard. Each action generates a counter-reaction, maintaining the universe's balance as per Newton's third law.
Forces in Motion
Forces are central to understanding motion. They influence the way objects move and interact with each other.

When an object is in motion or at rest, it is all due to the forces applied to it. Newton's Third Law of motion connects these forces together, ensuring that for every motion we observe, there's a cause behind it.

For instance, when the girl's head exerts a force on her neck during landing, this force sets her neck into a brief movement, though the outcome may not always be visibly noticeable. The reaction from her neck is what stabilizes her head from continuous movement. This reciprocal action ensures she doesn't continue to plunge further into motion. It is these unseen but deeply connected forces that play a crucial role in controlling how and why objects behave the way they do.
Equal and Opposite Forces
The principle of equal and opposite forces is fundamental to many aspects of physics. Newton's Third Law not only lays the groundwork for understanding these interactions but also ensures that they remain predictable and consistent in nature.

Whenever a force is exerted, the world guarantees that another force of equal magnitude but opposite direction counteracts it. In the given exercise, the forces exerted between the girl's head and neck are prime examples of this. Her head's force against her neck is counterbalanced by an equal force from her neck; this is an automatic, inherent response of the physical world.
  • The force from the head: downward
  • The force from the neck: upward and equal
This balancing act ensures that for every push, there is a corresponding push back, maintaining equilibrium. This symmetry helps us understand not only stationary objects but also those in dynamic motion, confirming that nature follows a coherent set of rules. This concept allows us to predict and comprehend the outcomes of interactions in mechanical systems, from the complex orbits of planets to the simplicity of a ball bouncing off the ground.

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Most popular questions from this chapter

World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude \(15 \mathrm{~m} / \mathrm{s}^{2} .\) How much horizontal force must a \(55 \mathrm{~kg}\) sprinter exert on the starting blocks to produce this acceleration? Which object exerts the force that propels the sprinter: the blocks or the sprinter herself?

A \(68.5 \mathrm{~kg}\) skater moving initially at \(2.40 \mathrm{~m} / \mathrm{s}\) on rough horizontal ice comes to rest uniformly in \(3.52 \mathrm{~s}\) due to friction from the ice. What force does friction exert on the skater?

A \(5.60 \mathrm{~kg}\) bucket of water is accelerated upward by a cord of negligible mass whose breaking strength is \(75.0 \mathrm{~N}\). If the bucket starts from rest, what is the minimum time required to raise the bucket a vertical distance of \(12.0 \mathrm{~m}\) without breaking the cord?

A student of mass \(45 \mathrm{~kg}\) jumps off a high diving board. What is the acceleration of the earth toward her as she accelerates toward the earth with an acceleration of \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) ? Use \(6.0 \times 10^{24} \mathrm{~kg}\) for the mass of the earth, and assume that the net force on the earth is the force of gravity she exerts on it.

A man is dragging a trunk up the loading ramp of a mover's truck. The ramp has a slope angle of \(20.0^{\circ},\) and the man pulls upward with a force \(\vec{F}\) whose direction makes an angle of \(30.0^{\circ}\) with the ramp (Fig. E4.4). (a) How large a force \(\vec{F}\) is necessary for the component \(F_{x}\) parallel to the ramp to be \(90.0 \mathrm{~N} ?\) (b) How large will the component \(F_{y}\) perpendicular to the ramp be then?

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