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

Kathy Williams

Kathy Williams

Answered question

2022-01-09

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.

Answer & Explanation

Mary Nicholson

Mary Nicholson

Beginner2022-01-10Added 38 answers

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
Wt=Wf=μmgl=KfalKitial=12m(vf2vi2)
l=12μg(vi2vf2)
=(120.259.8(321.652))m
=1.28m

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