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Chapter 6: Problem 121

A 46-kg person drinks 500 g of milk, which has a "caloric" value of approximately \(3.0 \mathrm{~kJ} / \mathrm{g}\). If only 17 percent of the energy in milk is converted to mechanical work, how high (in meters) can the person climb based on this energy intake? [Hint: The work done in ascending is given by \(m g h,\) where \(m\) is the mass (in kilograms), \(g\) the gravitational acceleration \(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right),\) and \(h\) the height (in meters).]

Short Answer

Expert verified
The person can climb to an approximately height of 565 meters based on the energy intake from the milk.

Step by step solution

01

Calculate the Total Energy in the Milk

The total energy contained in the milk can be calculated by taking the product of the mass of the milk and its energy density. The mass of the milk is given as 500g, and the energy density as \(3.0 \mathrm{~kJ/g}\). So, the total energy can be calculated using the formula: \(Energy =Mass \times Energy Density \rightarrow Energy =500g \times 3.0 kJ/g = 1500kJ \)
02

Calculate the Energy Converted to Mechanical Work

It's given that only 17 percent of the total energy in the milk can be converted to mechanical work. This means that the work energy will be: \(Work Energy =Total Energy \times Conv. Efficiency \rightarrow Work Energy = 1500kJ \times 0.17 = 255kJ\)
03

Calculate the Height

Now, we know that the work done in ascending is equal to \(mgh\), where \(m\) is the mass of the person, \(g\) is the gravitational acceleration, and \(h\) is the height. So, rearranging for \(h\), we get the formula: \(h = Work / (m \times g)\). Finally, substituting the known values (remembering to convert the work energy from kJ to J, since 1kj = 1000J), we find: \(h =255kJ \times 1000/(46kg \times 9.8m/s^2) = 565m\)

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