A 30-N iPad is dropped from a height of 10 m and strikes the ground with a speed of 13 m/s. What average force of air friction acted on the iPad as it fell?​

Initially the PE of the object was W * h = 30 * 10 = 300 Joules

The KE of the object when it struck the ground was 1/2 M v^2

KE = 1/2 * 30/9.8 * 13^2 = 259 J

So the object lost 41 J to friction during the fall

Since Work = Force * distance

Force = 41 J / 10 m = 4.1 N  (the average force of friction)

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