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While doing a chin-up, a man lifts his body 0.40 m. (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the \(total\) percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.) (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 J of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.

Short Answer

Expert verified
(a) 3.924 J/kg; (b) 5.61%; (c) 2.80%; (d) Shorter arms reduce work needed.

Step by step solution

01

Calculate the Work Done per Kilogram (Part a)

The work done by a force is calculated using the formula \( W = F \cdot d \), where \( F \) is the force and \( d \) is the distance moved in the direction of the force. Since the man lifts his body a distance \( d = 0.40 \) m against gravity, the force is equal to his mass times the acceleration due to gravity \( g = 9.81 \) m/sĀ². The work done per kilogram of body mass is \( W = m \cdot g \cdot d / m = g \cdot d = 9.81 \times 0.40 = 3.924 \) J/kg.
02

Determine the Percentage of Muscle Mass (Part b)

The muscles can generate 70 J of work per kilogram of muscle mass. Thus, the work required to lift the man's body for a chin-up is 3.924 J/kg (as calculated in Step 1) per kilogram of his total body mass. This means the required work is 3.924 J for each kilogram of muscle mass. Therefore, the percentage of his body's mass that these muscles constitute is \( \frac{3.924}{70} \times 100 \approx 5.61\% \).
03

Adjust Calculation for Son (Part c)

The son's arms are half as long, so the distance traveled during a chin-up is 0.20 m. Recalculate the work done per kilogram as \( W = g \cdot d = 9.81 \times 0.20 = 1.962 \) J/kg. For the son's muscles working at 70 J/kg, the percentage of muscle mass is \( \frac{1.962}{70} \times 100 \approx 2.80\% \).
04

Explain Muscle Efficiency in Children vs. Adults (Part d)

Children may find chin-ups easier than adults partly because they are generally lifted a shorter vertical distance due to their shorter arms. As a result, they do less work for the same motion, and even though their muscle mass ratio might be similar, the required work capacity per kilogram is less due to the reduced distance. This means that even if the percentage of muscle mass is the same, the efficiency required is less.

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Key Concepts

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

Chin-up Exercise
A chin-up exercise is a popular strength-training activity that primarily targets the upper body muscles. In a chin-up, an individual pulls themselves up using their arms until their chin is above a horizontal bar. This movement requires significant strength from various muscle groups, including the biceps, back, and shoulders. The simplicity of the equipment ā€“ just a bar ā€“ makes chin-ups a favorite in both gyms and home workouts.

Chin-ups not only build muscle but also improve endurance and grip strength. They engage the body's central and upper regions, demanding coordination and power. For many, achieving a chin-up is a feat of overcoming gravity, leveraging one's muscle power to pull one's own weight vertically. This exercise thus forms a fundamental component of workouts aimed at enhancing total body strength.
Muscle Mass Percentage
Muscle mass percentage refers to the proportion of a person's body mass that is made up of muscles. This metric is crucial in understanding an individual's physical fitness and metabolic health. Typically, muscle mass makes up about 40% of body weight in healthy adults, but this can vary based on factors like age, gender, and fitness level.

In the context of chin-ups, the percentage of muscle mass defines how efficiently a person can perform the exercise. Muscles act as the engine for movement ā€“ the more muscle mass relative to body weight, the potentially more power the individual can exert. This becomes particularly relevant when comparing performance across demographics, like adults versus children, or comparing two individuals with different body compositions.
Work Per Kilogram
The concept of work, especially "work per kilogram," is crucial in understanding physical exertion during exercise. Work, in physics, is defined as the energy transferred by a force acting through a distance. The formula used is: \[ W = F \cdot d \] where \( W \) is work, \( F \) is the force, and \( d \) is the distance.

In chin-ups, force is exerted against gravity to lift the body's weight. The work done is measured per kilogram to standardize performance across different body sizes. For example, if lifting a body 0.40 m requires 3.924 J/kg, it shows the energy spent per kilogram. This kind of measurement helps compare how intensely different bodies work under similar conditions, revealing efficiency and muscle capability.
Gravity and Forces
Gravity is a fundamental force in physics that affects every activity here on Earth. It imparts a constant downward force on objects, proportional to their mass, and is typically quantified as \( 9.81 \text{ m/s}^2 \). In exercises like the chin-up, gravity plays an essential role as it is the force one has to overcome to lift their body vertically.

Understanding forces is vital to evaluating exercises as it involves analyzing how much force muscles need to exert to counteract gravity. This becomes important in calculating the "work" performed during physical activities. When performing a chin-up, the force required to lift must equal the gravitational force acting on one's mass. This knowledge lays the groundwork for improving performance, ensuring safe exercise practices, and designing workout regimens that meet personal fitness goals.

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Most popular questions from this chapter

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point \(P\), it encounters a rough section. The coefficient of friction there is not constant; it starts at 0.100 at \(P\) and increases linearly with distance past \(P\), reaching a value of 0.600 at 12.5 m past point \(P\). (a) Use the work\(-\)energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of 0.100?

Use the work\(-\)energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0-m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 m into the air. How fast was the boulder moving just as it left the volcano?

You are a member of an Alpine Rescue Team. You must project a box of supplies up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_k\). Use the work\(-\) energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g\), \(h\), \(\mu_k\), and \(\alpha\).

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On a farm, you are pushing on a stubborn pig with a constant horizontal force with magnitude 30.0 N and direction 37.0\(^\circ\) counterclockwise from the +\(x\)-axis. How much work does this force do during a displacement of the pig that is (a) \(\overrightarrow{s} = (5.00 \, \mathrm{m})\hat{\imath}\); (b) \(\overrightarrow{s} = - (6.00 \, \mathrm{m})\hat{\jmath}\); (c) \(\overrightarrow{s} = - (2.00 \, \mathrm{m})\hat{\jmath} + (4.00 \, \mathrm{m}) \hat{\jmath}\)?

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