Posted by **help** on Sunday, September 28, 2008 at 6:09pm.

A 68.5kg skater moving initially at 2.40m/s on rough horizontal ice comes to rest uniformly in 3.52s due to friction from the ice. I need help finding the force friction exert on the skater.

- physics -
**drwls**, Sunday, September 28, 2008 at 6:22pm
Use the rule that impulse = momentum change.

Friction force * (time applied) = loss of linear momentium

F * 3.52 s = (68.5 kg)*(2.40 m/s)

Solve for F, which will be in Newtons

- physics -
**help**, Sunday, September 28, 2008 at 6:30pm
Thank you

- physics -
**DR. **, Wednesday, September 24, 2014 at 7:57pm
Use average acceleration equation to find the acceleration: a=delta(v)/delta(t), then plug into the newtons 2nd law of motion, which is the equation vector form.

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