You\'re now moving a chair with a mass of 85.0 kg in the same way as the previou

Question

You're now moving a chair with a mass of 85.0 kg in the same way as the previous problem, but part way down the sidewalk the ice has melted. To account for the frictional force you know will now be present, you push against the chair a little harder, with a force of 63.0 N. Just as you reach the de-iced part of the sidewalk, you and the chair are moving at 2.60 m/s, but despite your best efforts the chair quickly slows down and comes to rest within 5.4 s. What is the magnitude of the frictional force between the chair and the sidewalk?

Explanation / Answer

F=63N, frictional force fk =?

F-fk =m*a

where a=(v2-v1)/t =(0-2.6)/5.4 =-0.5m/s^2

=>63-fk =-85*0.5 =>fk =63+85/2 =105.5N

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