On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed to 1.65 m/s due to a friction force that is 25% of her weight. Use the work–energy theorem to find the length of this rough patch.

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Mary Nicholson · Accepted Answer

Step 1    The ice-skater experiences three forces during his/her transition across the rough patch: gravity, the normal force, and the force of friction. The first two cancel out, so that the force of friction is the total force acting on the skater, equal to Ff=μmg. Since Ffmg=14 , we have μ=14. The friction is opposite to the displacement and so the work done by friction is Wf=−μmgl, where l is the length of the rough patch.        Now, if the initial and the final speeds of the skater were vi=3ms and vf=1.65, respectively, by the work energy theorem we have    W→t=Wf=−μmgl=Kf∈al−K∈itial=12m(vf2−vi2)⇒    l=12μg(vi2−vf2)    =(12⋅0.25⋅9.8(32−1.652))m    =1.28m

On an essentially frictionless, horizontal ice rink, a skater moving

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