Question

d10 ocean Starting from rest, a penguin weighing 4kg, slides down a snowy incline at 66.8° to horizontal. The incline is 21.75m long (d 1 in diagram). Then the penguin slides along a long snow-covered horizontal section of length 18.8m (d2 in diagram), then off of a 33.18m tall cliff (h in diagram) into the ocean. The coefficient of friction between the penguin and all snowy surfaces is 0.266. Ignoring air drag, what is the penguin's speed when reaching the water surface? Include SI units. Report your answer to 3 significant digits.

Answer 1

Mycosa & mg ingsing Let a be the accleration of the perquin in the inclined plane. Here, So, mgsina - umgeoso ama, 0 - 66.80 or a ag(sina - ucose) di = 21175 m M = 0.266 =908 sin 66'8° - 0:266 cos 66'8") = 7.981 m/s . At the end of inclined plane, the velocity is 9 = √ 2a, d2 = 12x7°981x21175 - 18'633 m/s for the horizontal portion. let the acceleration be az. umg im Fama,cosa - 02 . = 66.8" d₂ = 1818 m. So: ma, coso - ung = ma₂ or, O2 = a, coso - ug = 7.981 cos 66'8° - 0' 26698 0.537 m/s. Velocity at the end of horizontal portion , 10% = L07 - 202d2 - 18.6337 - 2X0153781818 or U₂ = 18.083 mle.

water surface, the relocity Before reaching the of the penguin Here, it falls under 8 = 0 +29h - 18.083 + 2x9.8x33:18 07,63 = 31 262 m/s h=33:18 m