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In climbing up a rope, a 62-kg athlete climbs a vertical distance of 5.0 m in 9.0 s. What minimum power output was used to accomplish this feat?

Short Answer

Expert verified
The minimum power output is approximately 338 Watts.

Step by step solution

01

Determine the Work Done

The work done by the athlete can be calculated using the formula for gravitational work, which is given by \( W = mgh \), where \( m \) is the mass of the athlete, \( g \) is the acceleration due to gravity, and \( h \) is the height climbed. Given \( m = 62 \) kg, \( g = 9.8 \) m/s², and \( h = 5.0 \) m, the work done is \( W = 62 \times 9.8 \times 5 = 3038 \) Joules.
02

Calculate the Power Output

Power is the rate at which work is done, calculated using \( P = \frac{W}{t} \), where \( W \) is the work done and \( t \) is the time taken. We found that \( W = 3038 \) Joules and \( t = 9.0 \) seconds. Therefore, the power output is \( P = \frac{3038}{9} \approx 337.56 \) Watts.

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Key Concepts

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

Gravitational Work Calculation
Work in physics is the measure of energy transfer that occurs when an object is moved over a distance by an external force. In the context of gravitational work, this involves lifting an object against the force of gravity. To calculate gravitational work, we use the formula: \[ W = mgh \] where:
  • \( W \) represents the work done in Joules (J).
  • \( m \) is the mass of the object in kilograms (kg).
  • \( g \) is the acceleration due to gravity, which on Earth is approximately 9.8 m/s².
  • \( h \) is the height in meters (m) over which the object is lifted.
The calculation of work involves understanding these concepts: an increase in the height \( h \), the object's mass \( m \), and the gravitational pull \( g \) all affect how much work is being done. In our example, the athlete climbs a 5-meter rope, requiring gravitational work of \( W = 62 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 5 \, \text{m} = 3038 \, \text{J} \). This formula shows how energy is needed to overcome the gravitational force acting on the athlete's body.
Power Output Calculation
Power in physics quantifies how quickly work is done or energy is transferred. It tells us both how much work is done and how fast it is performed. The power output can be calculated using the formula: \[ P = \frac{W}{t} \], where:
  • \( P \) is the power output in Watts (W).
  • \( W \) is the work done in Joules (J).
  • \( t \) is the time in seconds (s) over which the work is done.
In our scenario, the athlete climbs a height of 5 meters in 9 seconds, performing 3038 Joules of work. Hence, the power output is computed by dividing the work done by the time taken, \( P = \frac{3038}{9} \approx 337.56 \text{ W} \). Understanding power is crucial for assessing energy efficiency and the ability to sustain a physical activity, providing insight into how the athlete uses their strength over time.
Mass and Gravity in Physics
Mass and gravity are foundational concepts in physics that determine how much force is needed to move an object and how that object behaves when subject to gravitational pull. **Mass** is a measure of the amount of matter in an object, typically measured in kilograms (kg). It remains constant regardless of location. On the other hand, **gravity** is a force that pulls objects towards each other, with a significant example on Earth being the pull toward its center.
The gravitational force acting on an object can be expressed as \( F = mg \), where \( F \) is the force in Newtons (N), \( m \) is mass, and \( g \) represents the acceleration due to gravity, approximately 9.8 m/s² on Earth. This force determines how much energy is required to lift an object against gravity.
For the climbing athlete, the mass was 62 kg, meaning the gravitational force acting on them was \( 62 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 607.6 \, \text{N} \). The greater the mass, the more force—and, consequently, energy—is required for movement. Mass and gravity together are crucial for understanding how easily objects can be lifted or moved in various physical situations.

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Most popular questions from this chapter

(II) The Lunar Module could make a safe landing if its vertical velocity at impact is 3.0 m/s or less. Suppose that you want to determine the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is (\(a\)) zero; (\(b\)) 2.0 m/s downward; (\(c\)) 2.0 m/s upward. Use conservation of energy to determine h in each case. The acceleration due to gravity at the surface of the Moon is 1.62 m/s\(^2\).

(II) A spring with \(k =\) 83 N/m hangs vertically next to a ruler. The end of the spring is next to the 15-cm mark on the ruler. If a 2.5-kg mass is now attached to the end of the spring, and the mass is allowed to fall, where will the end of the spring line up with the ruler marks when the mass is at its lowest position?

(III) Early test flights for the space shuttle used a "glider" (mass of 980 kg including pilot). After a horizontal launch at 480 km/h at a height of 3500 m, the glider eventually landed at a speed of 210 km/h. (\(a\)) What would its landing speed have been in the absence of air resistance? (\(b\))What was the average force of air resistance exerted on it if it came in at a constant glide angle of 12\(^\circ\) to the Earth's surface?

An airplane pilot fell 370 m after jumping from an aircraft without his parachute opening. He landed in a snowbank, creating a crater 1.1 m deep, but survived with only minor injuries. Assuming the pilot's mass was 88 kg and his speed at impact was 45 m/s,estimate: (\(a\)) the work done by the snow in bringing him to rest; (\(b\)) the average force exerted on him by the snow to stop him; and (\(c\)) the work done on him by air resistance as he fell. Model him as a particle.

(II) A baseball (\(m =\) 145 g) traveling 32 m/s moves a fielder's glove backward 25 cm when the ball is caught.What was the average force exerted by the ball on the glove?

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