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While running, a person dissipates about \(0.600 \mathrm{J}\) of mechanical energy per step per kilogram of body mass. If a 60.0 -kg runner dissipates a power of \(70.0 \mathrm{W}\) during a race, how fast is the person running? Assume a running step is \(1.50 \mathrm{m}\) long.

Short Answer

Expert verified
The speed of the person is \(10.5 km/hour\).

Step by step solution

01

Calculate the total energy dissipated per step

The person dissipates \(0.6J\) per step per kilogram of body mass. So, for a 60-kg person, energy dissipated per step is \(0.6 J/kg * 60 kg = 36 J/step\).
02

Compute steps in one second

Given that the person dissipates 70W power, and power is defined as work done per time or energy transferred per time. Therefore, \(70 joules/sec = 36 joules/step \Rightarrow steps/sec = 70 / 36 = 1.944 steps/sec\).
03

Determine the speed of the person

The speed of the person = steps/sec * distance/step = \(1.944 steps/sec * 1.5 meter/step = 2.916 m/sec\).
04

Convert the speed from meter/sec to km/hour

As 1 meter/sec = 3.6 km/hour, then the speed of the person = \(2.916 m/sec * 3.6 km/hour/m = 10.5 km/h\).

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Key Concepts

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

Work-Energy Principle
Understanding the work-energy principle is crucial when analyzing how objects move and how different forms of energy are converted. In physics, this principle states that the work done on an object by external forces equals the change in its kinetic energy. Work is the product of force and displacement in the direction of the force, and kinetic energy is a form of energy related to motion.

For example, if our runner pushes against the ground to move forward, the work done by their muscles translates into kinetic energy, allowing them to gain speed. In the running person scenario, each step dissipates energy due to friction and air resistance, this dissipation being a form of negative work that reduces the runner's mechanical energy. The work-energy principle helps us quantify this energy loss and ultimately find out how it affects the runner's speed.
Power in Physics
Power is a crucial concept that deals with the rate at which work is done or energy is transferred. In physics, it is defined as work done per unit of time. The unit of power is the watt (W), where one watt equals one joule per second.

When our hypothetical runner dissipates a power of 70.0 W during their race, they're effectively losing 70 joules of energy every second due to forces like friction and air resistance. Understanding power allows us to calculate how quickly they must be moving to sustain this energy loss, which leads to the calculation of steps taken per second. This step-by-step energy analysis using power reveals not just the dissipation rates but also the intensiveness of the runner's activity in real-time.
Converting Units of Speed
Being adept at converting units of speed is essential in physics to ensure clarity and precision in communication and calculations. Speed is typically measured in meters per second (m/s) in the scientific community, but for everyday use, kilometers per hour (km/h) is more common. The conversion is straightforward: 1 m/s equals 3.6 km/h.

To visualize this, imagine traveling at a speed of 1 meter every second; in one hour, you would have covered 3,600 meters, or 3.6 kilometers. Therefore, by multiplying the speed in m/s by 3.6, you get the speed in km/h. In our runner's case, their speed was calculated as 2.916 m/s. When converted, this is equivalent to 10.5 km/h (2.916 m/s times 3.6 km/h per m/s). This final conversion makes it easier to understand the runner's pace in a more commonly used metric, especially useful for real-world applications such as sporting events.

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Most popular questions from this chapter

A 15.0 -kg block is dragged over a rough, horizontal surface by a \(70.0-\mathrm{N}\) force acting at \(20.0^{\circ}\) above the horizontal. The block is displaced \(5.00 \mathrm{m},\) and the coefficient of \(\mathrm{ki}-\) netic friction is \(0.300 .\) Find the work done on the block by (a) the 70 -N force, (b) the normal force, and c) the gravitational force. (d) What is the increase in internal energy of the block-surface system due to friction? (e) Find the total change in the block's kinetic energy.

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The spring constant of an automotive suspension spring increases with increasing load due to a spring coil that is widest at the bottom, smoothly tapering to a smaller diameter near the top. The result is a softer ride on normal road surfaces from the narrower coils, but the car does not bottom out on bumps because when the upper coils collapse, they leave the stiffer coils near the bottom to absorb the load. For a tapered spiral spring that compresses \(12.9 \mathrm{cm}\) with a \(1000-\mathrm{N}\) load and \(31.5 \mathrm{cm}\) with a \(5000-\mathrm{N}\) load, (a) evaluate the constants \(a\) and \(b\) in the empirical equation \(F=a x^{b}\) and \((b)\) find the work needed to compress the spring \(25.0 \mathrm{cm}.\)

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