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.

Question

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.

drwls · Answer

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

help · Answer

Thank you

DR. · Answer

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.