A 65-kg hiker climbs to the top of a mountain 4200 m high....

A 65-kg hiker climbs to the top of a mountain 4200 m high. The climb is made in 4.6 h starting at an elevation of 2800 m. Calculate ($a$) the work done by the hiker against gravity, ($b$) the average power output in watts and in horsepower, and ($c$) assuming the body is 15% efficient, what rate of energy input was required.

Key Concepts

Energy Efficiency

Energy efficiency refers to the fraction of the input energy that is successfully converted into useful output work, while the remainder is lost to waste, such as heat. In biomechanical contexts, such as human activity, this concept highlights that the body’s metabolic processes are not 100% efficient, and determining the rate of energy input requires accounting for these inefficiencies.

Gravitational Potential Energy

This concept refers to the energy possessed by an object due to its position in a gravitational field. The energy is calculated using the formula mgh, where m is the mass, g is the acceleration due to gravity, and h is the height difference. It represents the work done against gravity to elevate an object to a higher altitude and is a central idea in energy conservation problems in mechanics.

Work

Work is defined as the transfer of energy when a force is applied over a distance. In the context of climbing, the work done against gravity is equivalent to the gravitational potential energy gained. This concept emphasizes that work is the product of force and displacement in the direction of the force, serving as a fundamental aspect of energy transfer in mechanical systems.

Power

Power is the rate at which work is performed or energy is transferred over time. It is calculated by dividing the work done by the time taken to perform that work. In practical applications, power is measured in watts, which quantifies how quickly energy is expended or converted.

Unit Conversion

The process of unit conversion involves translating a measurement from one unit to another to ensure consistency and accuracy in calculations. In problems involving work and power, it is often necessary to convert time units (such as hours into seconds) or energy units (for instance, converting watts to horsepower), thereby facilitating proper quantitative analysis in different measurement systems.

Transcript

00:01 Okay, so when it workdown is equal to force times the distance.

00:04 So the force in this case is the gravity of the person, which is mg.

00:08 And the distance here is the high difference because the person is starting at an elevation of 2 ,800 meter.

00:18 Okay? so therefore, we know the mass for the hiker is 65 kilograms, and the acceleration of gravity is 9 .8 meter per second square.

00:26 And the high difference is the top mountain, minus 10%.

00:31 The elevation point, which is 4 ,200 meter minus 2 ,800 meter.

00:35 It gives us 1 ,400 meter.

00:37 So therefore, the workdown should be equal to 65 kilograms times 9 .8 meter per second square, and then times 1 ,400 meter.

00:55 And this will give us the workdown is about 8 .92 times 102 power 5, you and for the power, we know the power is equal to work down divided by the time interval.

01:15 So since it takes the person 4 .6 hour to climb to the top.

01:22 So therefore, we know 4 .6 hour is equal to 16 ,500 seconds because in one hour there are 3 ,600 seconds.

01:34 And we know work down from the last question.

01:36 So therefore, we determine the average power, which is equal to 8 .92 times 10 to power 5ju over 16 ,560 seconds.

01:52 And this will give us the power is about 53 .86 watts.

02:03 And for the last question, well, we know that the efficiency rates is 15%...

Please give Ace some feedback