An ice skater of mass m is given a shove on a frozen pond. After the shove she has a speed of v_1 = 2 m/s. Assume that the only horizontal force that acts on her is a slight frictional force between the blades of the skates and the ice.
Use the net work-kinetic energy theorem to find the distance the skater moves before coming to rest. Assume that the coefficient of kinetic friction between the blades of the skates and the ice is u^kin = 0.12.
2006-10-25 16:36:43 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Answer 1 gave a formula for distance d that is the height if all kinetic energy were converted to lifting the skater. Apparently the coefficient of friction u got forgotten. With u included we have mgud = 1/2 mv^2, resulting in d = v^2/(2gu).
2006-10-26 03:32:12 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
The work done by the skater on the ice through friction is Fd cos (theta) = -mgud since the force and motion are opposite. The energy available for this is the kinetic energy of the skater = 1/2 mv^2. So when the skater transfers all of this kinetic energy into the ice she will stop and her total final energy is zero so E - W = 0. So mgd = 1/2 mv^2 or gd = 1/2 v^2. Finally, d = v^2/(2g). You do the math.
2006-10-26 03:21:24 · answer #2 · answered by justaguy 2 · 0⤊ 0⤋
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2016-11-25 21:07:05 · answer #3 · answered by ? 4 · 0⤊ 0⤋