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Chapter 5: Problem 52

While running, a person dissipates about \(0.60 \mathrm{~J}\) of mechanical energy per step per kilogram of body mass. If a \(60-\mathrm{kg}\) person develops a power of \(70 \mathrm{~W}\) during a race, how fast is the person running? (Assume a running step is \(1.5 \mathrm{~m}\) long.)

Short Answer

Expert verified
The person is running at a speed of \(2.92 \mathrm{~m/s}\)

Step by step solution

01

Determine energy dissipated per step

The total energy dissipated per step is given by the energy dissipated per step per kilogram of body mass times the person's body mass. Hence, energy dissipated per step \(E_{step}\) is \(0.60 \mathrm{~J/kg} \times 60 \mathrm{~kg} = 36 \mathrm{~J}\) .
02

Calculate the time taken for each step

The power developed by the person, which is given as 70 W, is equal to the work done (or energy dissipated) per unit time. We can therefore find the time taken for each step \(t_{step}\) from the equation \(Power = \frac{Work}{Time}\) or \(70 \mathrm{~W} = \frac{36 \mathrm{~J}}{t_{step}}\). Solving for \(t_{step}\) we get \(t_{step} = \frac{36 \mathrm{~J}}{70 \mathrm{~W}} = 0.514 \mathrm{~s}.\)
03

Determine the speed of the person

The speed of the person can be obtained from the distance covered per unit time. Since the distance of each step is 1.5 m, the speed \(v\) can be calculated from \(v = \frac{Distance}{Time} = \frac{1.5 \mathrm{~m}}{0.514 \mathrm{~s}} = 2.92 \mathrm{~m/s}.\)

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Key Concepts

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

Understanding Power Output in Running
Power output in the context of running refers to the rate at which a person is using or dissipating energy. It's about how much energy is used during physical activity over an amount of time. In this exercise, we look at how the power output is utilized by a runner during a race. The power output provided is 70 Watts. Watts measure work done over time, with 1 Watt equal to 1 Joule per second.
To visualize: if a runner is using a constant rate of energy equating to 70 Joules every second, this is their power output. It's a measure of efficiency and exertion, telling us how energetically demanding the task is for the runner at any given moment.
  • Power (Watts) = Work done (Joules) / Time (seconds)
  • It's crucial to understand that high power output can signify either a high energy expenditure over a short period, or consistent energy use over a longer time span.
Calculating Energy per Step
Energy per step is a critical measure of how much energy is utilized with each stride a runner takes. For each kilogram of body mass, a runner dissipates 0.60 Joules of energy. This allows us to calculate the total energy per step, by multiplying this number by the runner's weight. For a 60 kg person, this means each step requires 36 Joules (since 0.60 J/kg x 60 kg = 36 J).
The energy per step gives insight into the runner's efficiency. A higher energy cost per step might indicate either more intense movement or less efficient mechanics.
  • This measurement helps in understanding the impact that body mass has on energy use during activities like running.
  • Regularly monitoring energy per step can assist athletes in optimizing performance by adjusting technique, stride length, or even shoe choice.
Exploring Running Speed
Running speed is determined by how quickly a runner covers a certain distance. It is calculated as the distance per step divided by the time per step. In our case, each step is 1.5 meters long, and the time per step previously calculated is 0.514 seconds. Thus, the speed comes out to 2.92 meters per second (1.5 m / 0.514 s).
Speed is a central component in many athletic endeavors. It reflects not only fitness but also how efficiently energy is converted to forward motion.
  • A focused look at running speed can help athletes evaluate their progress, tailor their training regimen, and set realistic goals.
  • Increasing running speed can be achieved by reducing the time per step or increasing the length of each step, while maintaining efficient energy use.

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Most popular questions from this chapter

The electric motor of a model train accelerates the train from rest to \(0.620 \mathrm{~m} / \mathrm{s}\) in \(21.0 \mathrm{~ms}\), The total mass of the train is \(875 \mathrm{~g} .\) Find the average power delivered to the train during its acceleration.

A raw egg can be dropped from a third-floor window and land on a foam-rubber pad on the ground without breaking. If a \(75.0\) -g egg is dropped from a window located \(32.0 \mathrm{~m}\) above the ground and a foam-rubber pad that is \(15.0 \mathrm{~cm}\) thick stops the egg in \(9.20 \mathrm{~ms}\), (a) by how much is the pad compressed? (b) What is the average force exerted on the egg after it strikes the pad? Note: Assume constant upward acceleration as the egg compresses the foam-rubber pad.

A sledge loaded with bricks has a total mass of \(18.0 \mathrm{~kg}\) and is pulled at constant speed by a rope inclined at \(20.0^{\circ}\) above the horizontal. The sledge moves a distance of \(20.0 \mathrm{~m}\) on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is \(0.500 .\) (a) What is the tension in the rope? (b) How much work is done by the rope on the sledge? (c) What is the mechanical energy Iost due to friction?

A 70 -kg diver steps off a \(10-\mathrm{m}\) tower and drops from rest straight down into the water. If he comes to rest \(5.0 \mathrm{~m}\) beneath the surface, determine the average resistive force exerted on him by the water.

A \(1.50 \times 10^{3}\) -kg car starts from rest and accelerates uniformly to \(18.0 \mathrm{~m} / \mathrm{s}\) in \(12.0 \mathrm{~s}\). Assume that air resistance remains constant at \(400 \mathrm{~N}\) during this time. Find (a) the average power developed by the engine and (b) the instantaneous power output of the engine at \(t=12.0 \mathrm{~s}\), just before the car stops accelerating.
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