The increase in potential energy is mgh = 75 x 9.8 x (490-250) = 176,400 J
The work the student does is 280,000 J
The efficiency is work out/work in = 176400/280000 = 63%
Michael W.
asked • 07/22/20A student climbs a hill from an elevation of 250 m to an elevation of 490 m. In doing so he expends 2.8 x 105J. The student’s mass is 75 kg. What is his climbing efficiency?
A student climbs a hill from an elevation of 250 m to an elevation of 490 m. In doing so he expends 2.8 x 105J. The student’s mass is 75 kg. What is his climbing efficiency?
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Michael W.
Thank you!07/22/20