A penguin slides at a constant velocity of 1.4 m/s down an icy incline. The incl

Question

A penguin slides at a constant velocity of 1.4 m/s down an icy incline. The  incline slopes above the horizontal at an angle of 6.9?degrees. At the bottom of the incline, the penguin slides onto a horizontal patch of ice. The coefficient of kinetic friction between the penguin and the ice is the same for the incline as for the horizontal slope. How much time is required for the penguin to slide to a halt after entering the horizontal patch of ice? A penguin slides at a constant velocity of 1.4 m/s down an icy incline. The  incline slopes above the horizontal at an angle of 6.9?degrees. At the bottom of the incline, the penguin slides onto a horizontal patch of ice. The coefficient of kinetic friction between the penguin and the ice is the same for the incline as for the horizontal slope. How much time is required for the penguin to slide to a halt after entering the horizontal patch of ice?

Explanation / Answer

force on the incline = g sin teta - umg cos theta

since the penguin slides at a cnstant velocity

force = 0

mg sin teta - umg cos theta = 0

this implies that u = tan theta

deacceleration n the flat srface = ug = tan theta * g

v= u + at

0 = 1.4 + (-ug)t

t = 1.4 / (ug) = 1.4 /(g tan theta) = 1.4 /(9.8 * tan6.9) = 1.18s

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