Power. 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 / sin  = (50 m)

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Presentation transcript:

Power

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 / sin  = (50 m) / sin 30° = 100 mDistance d = h / sin  = (50 m) / sin 30° = 100 m Work done by gravity W =  mg d sin  =Work done by gravity W =  mg d sin  =  (60 kg)(9.8 m/s2)(50 m) =  30 kJ  (60 kg)(9.8 m/s2)(50 m) =  30 kJ Gravity does negative work on the hiker d = 100 m -mg

Rate of Work  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 st = (50 m)/sin30°/(1/ms) =100 s W = 30 kJW = 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. Hiker does positive work to overcome gravity d = 100 m

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 m 2 / s 31 watt = 1 J/s = 1 N m/s = 1 kg m 2 / s 3  Energy can be measured in watt-seconds = joules.

Average Power  The walker had an average power output based on the work compared to the time. P = W / tP = W / t P = 30 kJ / 100 s = 300 WP = 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.

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

Power Plants  Electrical power is measured in watts. 60 W light bulb60 W light bulb 1000 MW power plant1000 MW power plant  The energy used is measured in power times time.  If electricity costs $ per kWh, how much does it cost to leave a a 1500 W floodlamp on all year? Energy used is W = Pt = (1500 W)(3.2 x 10 7 s) = 4.8 x J Cost is (4.8 x W s) * (1 kW / 1000 W) * (1 h / 3600 s)*(0.083 / kWh) = $1,100

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