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A man of mass \(80 \mathrm{kg}\) runs up a flight of stairs \(20 \mathrm{m}\) high in 10 s. (a) how much power is used to lift the man? \((b)\) If the man's body is \(25 \%\) efficient, how much power does he expend?

Short Answer

Expert verified
(a) The power used to lift the man is \(1,572.8 W\). (b) The power expended by the man, considering his body is 25% efficient, is \(6,291.2 W\).

Step by step solution

01

Determine the work done on the man

To calculate the work done, we will first find the gravitational potential energy gained by the man. The formula for gravitational potential energy is: \[U = mgh\] Where: \(U\) = gravitational potential energy \(m\) = mass of the man (\(80 kg\)) \(g\) = acceleration due to gravity (\(9.81 m/s^2\)) \(h\) = height of the stairs (\(20 m\)) Plug in the given values to find the gravitational potential energy.
02

Calculate the gravitational potential energy

\(U = (80 kg)(9.81 m/s^2)(20 m)\) \(U = 15,728 J\) The work done on the man is equal to the gravitational potential energy gained. Therefore, the work done is \(15,728 J\).
03

Calculate the power used to lift the man

The formula for power is: \[P = \frac{W}{\Delta t}\] where: - \(P\) = power - \(W\) = work done (\(15,728 J\)) - \(\Delta t\) = time taken (\(10 s\)) Plug in the given values.
04

Find the power used to lift the man

\(P = \frac{15,728 J}{10 s}\) \(P = 1,572.8 W\) The power used to lift the man is \(1,572.8 W\).
05

Calculate the total power expended considering the efficiency

We know that the man's body is 25% efficient. The power expended by the man is the power required to lift the man divided by the efficiency: \[P_{expended} = \frac{P}{efficiency}\] where: - \(P_{expended}\) = power expended by the man - \(P\) = power used to lift the man (\(1,572.8 W\)) - efficiency = \(0.25\)
06

Find the power expended by the man

\(P_{expended} = \frac{1,572.8 W}{0.25}\) \(P_{expended} = 6,291.2 W\) The power expended by the man, considering his body is 25% efficient, is \(6,291.2 W\). So, (a) the power used to lift the man is \(1,572.8 W\), and (b) the power expended by the man is \(6,291.2 W\).

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Key Concepts

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

Work-Energy Theorem
Understanding the work-energy theorem is crucial for grasping how work and energy are related. In physics, this principle states that the work done by all the forces acting on an object is equal to the change in its kinetic energy. In the context of the exercise, however, we're looking at gravitational potential energy, which is a form of mechanical energy.

When the man runs up the stairs, his muscles do work against the force of gravity. According to the work-energy theorem, the energy transferred to the man by his muscles doing this work becomes his potential energy at the top of the stairs. This is why the gravitational potential energy formula, \[U = mgh\], is used to calculate the work done on the man.

Yet, in the real world, this energy doesn't just appear; it comes from the man's body converting chemical energy (from food) into mechanical energy. This is referential to how engines convert fuel into motion or how a battery provides energy for an electric device to work. Making these connections can help students understand energy conversion, a central concept in physics and other scientific fields.
Power Calculation in Physics
Power is the rate at which work is done or energy is transferred. It is a concept that helps us understand how quickly energy conversions or work take place. The formula used in the exercise is \[P = \frac{W}{\Delta t}\], where \(P\) stands for power, \(W\) for work done, and \(\Delta t\) for the time interval. It's measured in watts (W), where one watt is equal to one joule per second.

The man in the exercise exerted energy to go up the stairs, and we calculated the power used based on the work his body performed within the time taken to complete the task. When students see power calculations, it's beneficial to illustrate practical applications, such as comparing energy usage of household appliances, assessing athletic performances, or understanding how fast a car engine can do work.
Efficiency in Physics
Efficiency in physics is a measure of how well energy is converted from one form to another. It's a way to assess the effectiveness of a machine or process in using energy to do work. The efficiency is given as a percentage, representing the ratio of useful output energy to the input energy.

In the exercise, the man's body acts as a machine, converting chemical energy from food into mechanical energy to climb stairs. However, not all of this energy goes into lifting his body upwards; some is lost as heat due to metabolic processes and friction. The 25% efficiency means only a quarter of the input energy is used for climbing, with the remainder being 'wasted.' It's important to clarify to students that efficiency is a key factor in energy conservation and technology development—from internal combustion engines to renewable energy systems like solar panels.

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Most popular questions from this chapter

A shopper pushes a grocery cart 20.0 \(\mathrm{m}\) at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction \(25.0^{\circ}\) below the horizontal. (a) What is the work done on the cart by friction? (b) What is the work done on the cart by the gravitational force? (c) What is the work done on the cart by the shopper? (d) Find the force the shopper exerts, using energy considerations. (e) What is the total work done on the cart?

Calculate the power output needed for a \(950-\mathrm{kg}\) car to climb a \(2.00^{\circ}\) slope at a constant \(30.0 \mathrm{m} / \mathrm{s}\) while encountering wind resistance and friction totaling \(600 \mathrm{N}\)

Consider a linear spring, as in Figure \(7.7(\) a), with mass \(M\) uniformly distributed along its length. The left end of the spring is fixed, but the right end, at the equilibrium position \(x=0, \quad\) is moving with speed \(v\) in the \(x\) -direction. What is the total kinetic energy of the spring? (Hint: First express the kinetic energy of an infinitesimal element of the spring \(d m\) in terms of the total mass, equilibrium length, speed of the right-hand end, and position along the spring; then integrate.)

A crate is being pushed across a rough floor surface. If no force is applied on the crate, the crate will slow down and come to a stop. If the crate of mass \(50 \mathrm{kg}\) moving at speed \(8 \mathrm{m} / \mathrm{s}\) comes to rest in 10 seconds, what is the rate at which the frictional force on the crate takes energy away from the crate?

One particle has mass \(m\) and a second particle has mass 2 \(m\). The second particle is moving with speed \(v\) and the first with speed 2 v. How do their kinetic energies compare?

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