/*! 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 67 The efficiency of a muscle durin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The efficiency of a muscle during weight lifting is equal to the work done in lifting the weight divided by the total energy output of the muscle (work done plus internal energy dissipated in the muscle). Determine the efficiency of a muscle that lifts a \(161-\mathrm{N}\) weight through a vertical displacement of \(0.577 \mathrm{m}\) and dissipates \(139 \mathrm{J}\) in the process.

Short Answer

Expert verified
Answer: The efficiency of the muscle during weight lifting is 40%.

Step by step solution

01

Find the work done by the muscle in lifting the weight

To determine the work done by the muscle in lifting the weight, we can use the formula: Work = Force × Displacement × cos(angle) Here, the force is the weight lifted (given as 161 N) and the displacement is the vertical distance the weight is lifted (given as 0.577 m). Since the force is acting in the same direction as the displacement, the angle between them is 0 degrees. The cosine of 0 degrees is 1. Work = 161 N × 0.577 m × 1 = 92.817 J The work done by the muscle in lifting the weight is 92.817 Joules.
02

Determine the total energy output of the muscle

Now, we need to find the total energy output of the muscle. This can be calculated by adding the work done in lifting the weight and the energy dissipated in the muscle during the process. The energy dissipated is given as 139 J. Total energy output = Work done in lifting the weight + Energy dissipated in the muscle Total energy output = 92.817 J + 139 J = 231.817 J The total energy output of the muscle is 231.817 Joules.
03

Calculate the efficiency of the muscle

With both the work done in lifting the weight and the total energy output of the muscle, we can calculate the efficiency using the formula: Efficiency = (Work done in lifting the weight) / (Total energy output of the muscle) Efficiency = (92.817 J) / (231.817 J) = 0.400 (approx) To express this as a percentage, multiply the value by 100. Efficiency = 0.400 × 100 = 40% The efficiency of the muscle during weight lifting is 40%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Within an insulated system, \(418.6 \mathrm{kJ}\) of heat is conducted through a copper rod from a hot reservoir at \(+200.0^{\circ} \mathrm{C}\) to a cold reservoir at \(+100.0^{\circ} \mathrm{C} .\) (The reservoirs are so big that this heat exchange does not change their temperatures appreciably.) What is the net change in entropy of the system, in \(\mathrm{kJ} / \mathrm{K} ?\)
On a cold day, Ming rubs her hands together to warm them up. She presses her hands together with a force of \(5.0 \mathrm{N} .\) Each time she rubs them back and forth they move a distance of \(16 \mathrm{cm}\) with a coefficient of kinetic friction of \(0.45 .\) Assuming no heat flow to the surroundings, after she has rubbed her hands back and forth eight times, by how much has the internal energy of her hands increased?
An inventor proposes a heat engine to propel a ship, using the temperature difference between the water at the surface and the water \(10 \mathrm{m}\) below the surface as the two reservoirs. If these temperatures are \(15.0^{\circ} \mathrm{C}\) and \(10.0^{\circ} \mathrm{C},\) respectively, what is the maximum possible efficiency of the engine?
An ideal gas is in contact with a heat reservoir so that it remains at a constant temperature of \(300.0 \mathrm{K}\). The gas is compressed from a volume of \(24.0 \mathrm{L}\) to a volume of 14.0 L. During the process, the mechanical device pushing the piston to compress the gas is found to expend \(5.00 \mathrm{kJ}\) of energy. How much heat flows between the heat reservoir and the gas and in what direction does the heat flow occur?
(a) Calculate the efficiency of a reversible engine that operates between the temperatures \(600.0^{\circ} \mathrm{C}\) and \(300.0^{\circ} \mathrm{C} .\) (b) If the engine absorbs \(420.0 \mathrm{kJ}\) of heat from the hot reservoir, how much does it exhaust to the cold reservoir?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.