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Safari File Edit View History Bookmarks Window Help 73% D. Tue Nov 10 20 AM E Session masteringphysics.com CH 11 HW Problem 11.46 Resources previous l 2 of 14 l next Problem 11.46 Part A A 68 kg hiker walks at 5.0 km/h up a 8% slope. The indicated incline is the ratio of the vertical distance and the horizontal distance expressed as percentage What is the necessary metabolic power? Hint: You can model her power needs as the sum of the 380 W power to walk on level ground plus the power needed to raise her body by the appropriate amount. Assume that the efficiency of the body in using energy is 25%. Express your answer to two significant figures and include the appropriate units. P. Value Units Submit My Answers Give Up Incorrect; Try Again Provide Feedback Continue

Explanation / Answer

Power = Energy ÷ time

As the hiker walks up the slope, the hiker’s potential energy increases.
PE = 68 * 9.8 * h = 666.4 * h
Since the hiker is moving up a 8% slope, h = 0.08 * distance up the slope.

Convert from km/h = m/s
1 km = 1000 m, 1 h 3600s
1 km/h = 1000/3600 = 10/36 = 5/18 m/s
5 km/h = 25/18 m/s 1.3889 m/s
Each second, the hiker moves approximately 1.3889 meters up the slope.

h = 0.08 * 25/18
PE = 666.4 * 0.08 * 25/18
This is approximately 74.04 J
Each second, the hiker’s potential energy increases this amount.
Power = Energy ÷ time, time = 1 second
So, this is her power.

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