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classmate Date Page Ans: Pastas About Helicoptes- Taking upwards disection downward direction as Negative as positive and Mass of Helicopter M = 760 kg Initial velocity of Helicoptes ū= oms. Velowly after 5 Time taken in Velocity change t = 5s Initial Height of Helicoptes His Om Eimal Height of Helicoptes Hf = 8.9m ✓ = + lo mls. 2 Now - Initial Kinetic Energy of Helicoptes kis_mu Ki= OJ I U= 0) Final Kinetic Energy of Helicoptes K= LMV2 Kf = LX760 X(102 = 38000J Hence- Change in kinetic Energy of Helicoptes AK= Kf-Ki = 38000 J D Whem, the State of Helicopter changes, from Stationary on ground to moving upwards.
Two fosceč act on the Helicoptes. im Vestically downwasd direction. (2) FO Lift force Vestically upward disection Now - Work done by Mg = engel - Mg (Hf -Hi) (1 mg Work done by Mg = -760X9.8 x 8.9 =-66287:23 Now Let accume that work done W. Then According to wolke energy us by E is
dussmate continued Date Page - Total Work done by mg + N Total Work done by f = Ak OR 66 297.3 +W: 38000 W = (38000 + 66287.9) We 104287.9J. This work has been done Interval At = 55.
A time ton Hence Formula By the Average Power = What Avesage Power by lift fosce = 104287.2 J Average Power by lift fosce = 20857:44 Watt Ana 5 Past (b): About Cyclists - Mass of Bicyclist M = 75 kq Avesage Power gemesated Park 615 W/ kg Distance covered in lace D3 187 km = 187000m. Average speed during Race Y= 11.8 mls. Calculation of Past (a): Power generated fos total mass P = Avešage Powed genered, Pang x Mass, M Hence P = 6.5x75 = 487.5 W Let at be the time taken to complete the race them Vat =D --
continued Hence <= 187000 S At = 16696.42865. Now let W be the total wolt dome then. P = W W = PAT At Hence- W=(487.5 X 166.96.4286) J W = 8139508.9425 J Hence- Wosk dome by Cyclist = 8139503.94253 Ams.
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Calculation of Pestib): Given that IJ = 2.389 X104 Nutsitional calories Hence- 8139508.9485J = 8139508.94.95 9.389x10-4 Nutritional Calosies Hence W = 1944.5287 Nutritional Calasies. Wosk done by Cyclist = 1944.5287 Nutritional Calorier Ans
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