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A penguin slides at a constant velocity of 3.57 m/s down an icy incline. The incline slopes above the horizontal at an angle of 9.85°. 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 patch. How much time is required for the penguin to slide to a halt after entering the horizontal patch of ice?

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Answer:t=0.81 s

Step-by-step explanation:

Given

Penguin slides down with constant velocity of 3.57 m/s

as the Penguin Slides with constant velocity therefore
F_(net) is zero on Penguin


F_(net)=mg\sin \theta -f_r


f_r=friction Force


f_r=\mu mg


\mu =coefficient of Kinetic friction


mg\sin \theta =\mu mg


\tan \theta =\mu


\mu =\tan 9.85=0.45

after reaching on floor final velocity of penguin will be zero after time t

thus


v=u+at

here
a=\mu g


a=0.45* 9.8=4.41 (deceleration)


0=3.57-4.41* t


t=(3.57)/(4.41)=0.809


t=0.81 s

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