Asked by help
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.
Answers
Answered by
drwls
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
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
Answered by
help
Thank you
Answered by
DR.
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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