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Q102P

Expert-verifiedFound in: Page 211

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**The summit of Mount Everest is 8850 m **** ****above sea level. (a) How much energy would a 90 kg**** ****climber expand against the gravitational force on him in climbing to the summit from sea level? (b) How many candy bars, at 1.25 MJ**** ****per bar, would supply an energy equivalent to this? Your answer should suggest that work done against the gravitational force is a very small part of the energy expended in climbing a mountain.**

- The energy that the climber expands to work against the gravitational force is $7.8\times {10}^{6}\mathrm{J}$
- Number of candy bars required are 6 bar .

The mass of the climber is, m = 90 kg

The energy that the climber gets per bar, E = 1.25 MJ $\left(\frac{{10}^{6}\mathrm{MJ}}{1\mathrm{MJ}}\right)=1.25\times {10}^{6}\mathrm{J}$

The height of the summit of Mount Everest, h = 8850 m

The acceleration due to gravity is, $\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^{2}$

**The energy of the climber against the gravitational force from the sea level shows the results that there is a change in potential energy.**

Formula:

Change in potential energy, $\u2206\mathrm{PE}=\mathrm{mg}\left({\mathrm{h}}_{2}-{\mathrm{h}}_{1}\right)$ (1)

The work or energy expanded by the climber to go against the gravitational force to climb to the summit of the mountain is given using equation (1):

W = Change in potential energy

$=90\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}\times \left(8850\mathrm{m}-0\mathrm{m}\right)\phantom{\rule{0ex}{0ex}}=7.8\times {10}^{6}\mathrm{kg}.{\mathrm{m}}^{2}/{\mathrm{s}}^{2}\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^{2}/{\mathrm{s}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=7.8\times {10}^{6}\mathrm{J}$

Hence, the value of the energy is $7.8\times {10}^{6}\mathrm{J}$ .

The number of candy bars that are required by the climber to supply energy is given as:

$\mathrm{n}=\frac{7.8\times {10}^{6}\mathrm{J}}{1.25\times {10}^{6}\mathrm{J}/\mathrm{bar}}\phantom{\rule{0ex}{0ex}}=6.2\mathrm{bar}\phantom{\rule{0ex}{0ex}}\approx 6\mathrm{bar}$

Hence, the required candies are 6 bar .

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