Answer to Question #83747 in Classical Mechanics for Edouina Alice Foday

Question

Assume that a 75kg skater on level ice has built up her speed to 3km/hr. Given the respective static and kinetic friction of steel of ice =us=0.02 and uo=0.01. How far will she coast before sliding friction dissipates her energy? Does the distance of the coasting depend on the mass of the skater? Explain why

Answer 1

The friction force does negative work on the skater reducing her kinetic energy eventually to zero. Hence, to find the distance d of the coasting, one should equate the initial kinetic energy E of the skater and the absolute value of the work done by the friction force. We have E = ½ mv2, where m is the mass of the skater, and v = 3 km/hr = 0.83 m/s is her initial speed. The absolute value of the friction force during sliding is given by F = mgu0, where g = 9.8 m/s2 is the acceleration of gravity, and the absolute value of the work done by this force is A = Fd = mgu0d. Equating E and A, we have ½ mv2 = mgu0d, whence d = v2/(2gu0). The result does not depend on the mass of the skater because both the kinetic energy of the skater and the friction force are proportional to her mass, so that the mass just cancels in the equation. The result also does not depend on the static friction coefficient us because the skater is not static but is sliding in this process. Substituting the numerical values into the obtained equation for d, we get d = 3.5 m.