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In an iconic movie scene, Forrest Gump (https://openstaxcollege.org/I/21ForrGumpvid) runs around the country. If he is running at a constant speed of \(3 \mathrm{m} / \mathrm{s},\) would it take him more or less energy to run uphill or downhill and why?

Short Answer

Expert verified
It would take more energy to run uphill due to the added force of gravity working against you. Conversely, running downhill would require less energy as gravity acts as an assisting force.

Step by step solution

01

Understanding the physical principles

When running at a constant velocity, you must exert force to overcome friction and other resisting forces like air drag. In running uphill, gravity adds to these resisting forces; thus more energy must be expanded. In contrast, running downhill, gravity acts as an assisting force, reducing total energy use.
02

Understanding the role of gravity

Gravity will pull anything downhill, so running downhill will require less energy. However, while going uphill, the same force of gravity will work against you, making you exert more energy.
03

Discussing energy usage

As one exerts more force (work) to run uphill, more energy is used according to the work-energy theorem. Work is equal to force times distance, and the force required is greater going uphill due to the added force of gravity. Therefore, the work done and thus the energy used is greater while running uphill.

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Key Concepts

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

Physics of Motion
The physics of motion delves into how and why objects move. This forms the backbone of understanding the effect of different forces on objects in motion, such as a person running. When dealing with the physics of motion, we have to consider various aspects like speed, velocity, and acceleration. To keep this clear, speed is simply how fast an object is moving, while velocity is speed in a given direction. Acceleration, on the other hand, is a change in velocity, which can include speeding up, slowing down, or changing direction.

In our context with Forrest Gump running at a constant speed of \(3 \text{m/s}\), the net force applied by him to stay at this velocity is zero in accordance with Newton's first law of motion, assuming there's no change in speed or direction, which means he's in a state of equilibrium motion. However, the presence of forces such as gravity and friction impact this equilibrium, particularly when the running terrain changes.
Force and Energy Relationship
The relationship between force and energy is central to understanding how different actions require varying amounts of energy. Force is essentially a push or pull upon an object resulting from the object's interaction with another object. Whenever there's a force, there's also an energy change. Energy comes in many forms, but when discussing motion, we're mainly looking at kinetic energy (the energy of motion) and potential energy (the energy stored due to an object's position).

The work-energy theorem provides a clear link between the concepts of work (force applied over a distance) and energy. It states that work done on an object results in a change in its kinetic energy. Essentially, this means that the more force you have to exert to overcome resistance (like friction or gravity), the more work you do and the more energy you spend. Thus, taking the climb uphill for Forrest would require more force against gravity, thereby increasing the amount of work done and energy used.
Gravity's Effect on Motion
Gravity is a force that pulls objects toward each other, and on Earth, it gives us weight and affects every motion we make, particularly in the vertical direction. The gravitational force is what makes running uphill more strenuous - it's like an additional resistance that Forrest Gump has to overcome.

When running uphill, gravity works against you, increasing your weight effectively and the force you must exert to move forward. This results in more work and thus more energy consumption. Conversely, when moving downhill, gravity assists in the movement, reducing the need for energy expenditure, as it helps to pull you down the slope. This assistance from gravity means less work is required for motion, implying that less energy is expended. In summary, gravity can either be a friend or foe depending on the direction of motion with respect to the gravitational pull.

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Most popular questions from this chapter

Shown below is a box of mass \(m_{1}\) that sits on a frictionless incline at an angle above the horizontal \(\theta=30^{\circ} .\) This box is connected by a relatively massless string, over a frictionless pulley, and finally connected to a box at rest over the ledge, labeled \(m_{2} .\) If \(m_{1}\) and \(m_{2}\) are a height \(h\) above the ground and \(m_{2}>>m_{1}:\) (a) What is the initial gravitational potential energy of the system? (b) What is the final kinetic energy of the system?

A box is dropped onto a spring at its equilibrium position. The spring compresses with the box attached and comes to rest. since the spring is in the vertical position, does the change in the gravitational potential energy of the box while the spring is compressing need to be considered in this problem?

A body of mass \(m\) and negligible size starts from rest and slides down the surface of a frictionless solid sphere of radius \(R\). (See below.) Prove that the body leaves the sphere when \(\theta=\cos ^{-1}(2 / 3)\)

A block leaves a frictionless inclined surfarce horizontally after dropping off by a height \(h\). Find the horizontal distance \(D\) where it will land on the floor, in terms of \(h, H,\) and \(g\).

A particle of mass \(2.0 \mathrm{kg}\) moves under the influence of the force \(F(x)=\left(-5 x^{2}+7 x\right)\) N. Suppose a frictional force also acts on the particle. If the particle's speed when it starts at \(x=-4.0 \mathrm{m}\) is \(0.0 \mathrm{m} / \mathrm{s}\) and when it arrives at \(x=4.0 \mathrm{m}\) is \(9.0 \mathrm{m} / \mathrm{s},\) how much work is done on it by the frictional force between \(x=-4.0 \mathrm{m}\) and \(x=4.0 \mathrm{m} ?\)

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