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Understanding Work and Power in Uphill Hiking: Mechanics of Force and Energy

This article explores the work done against gravity during uphill hiking, focusing on a hiker with a mass of 60 kg ascending a 30° hill, resulting in a vertical height increase of 50 meters. We calculate the work done by gravity, the rate of work (power) during walking and running, and compare the energy output in watts. Through examples, we demonstrate how power varies with time and activity level. We also discuss the implications of power consumption in everyday life, revealing the costs associated with electrical energy usage based on power output.

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Understanding Work and Power in Uphill Hiking: Mechanics of Force and Energy

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Presentation Transcript

  1. Power

  2. Uphill • You (m = 60 kg) hike up a 30° hill with a net height increase (h = 50 m). What work is done by gravity? • Distance d = h / sinq = (50 m) / sin 30° = 100 m • Work done by gravity W = -mg d sinq = • -(60 kg)(9.8 m/s2)(50 m) = -30 kJ d = 100 m Gravity does negative work on the hiker -mg

  3. You (60 kg) walk up a 30° hill with a net height increase of 50 m at 1 m/s. t = (50 m)/sin30°/(1/ms) =100 s W = 30 kJ You run up the same 30° hill with a net height increase of 50 m at 4 m/s. t = 25 s, W = 30 kJ But, running is harder! The rate of work has increased by 4 times by running. Rate of Work Hiker does positive work to overcome gravity d = 100 m

  4. The Watt • The rate of work is called power. • The SI unit of power is the watt (W). • 1 watt = 1 J/s = 1 N m/s = 1 kg m2 / s3 • Energy can be measured in watt-seconds = joules.

  5. The walker had an average power output based on the work compared to the time. P = W / t P = 30 kJ / 100 s = 300 W The runner generated the same work in one quarter of the time. P = 30 kJ / 25 s = 1200 W When running seems harder, it isn’t work, it’s power. Average Power

  6. Instantaneous Power • Work does not have to be uniform over time. • Moving over a series of hills and valleys (changing work) • Walking and running (changing rate) • The power expended at one instant is the limit of work done over a very small time interval.

  7. Electrical power is measured in watts. 60 W light bulb 1000 MW power plant Energy used is measured in power times time. If electricity costs $ 0.083 per kWh, how much does it cost to leave a 1500 W floodlamp on all year? Energy used is W = Pt = (1500 W)(3.2 x 107 s) = 4.8 x 1010 J Cost is (4.8 x 1010 W s) * (1 kW / 1000 W) * (1 h / 3600 s)*(0.083 / kWh) = $1,100 Power Plants

  8. Power is work per time. Work is force acting over a distance Distance per time is velocity Power is force times velocity. Force and Velocity v F next

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