A skier traveling 11.0 m/s reaches the foot of a steady upward 19.6° incline and glides 13.8 m up along this slope before coming to rest. What was the average coefficient of friction?
3 answers
Use equation mgh + mgh(initial) = -1/2mv^2 - 1/2 mv(initial) + W
KE = PE + W_friction
1/2 * mv^2 = mgh + mu*mgcos(theta)d
----- m is cancel out
--- sin(theta) = heigh/ distance( inclide) ===> h = d*sin(theta)
1/2 v^2 = g (d*sin(theta)) + mu* (g*cos(theta))*d
now you can solve for mu ^^ Good luck.
1/2 * mv^2 = mgh + mu*mgcos(theta)d
----- m is cancel out
--- sin(theta) = heigh/ distance( inclide) ===> h = d*sin(theta)
1/2 v^2 = g (d*sin(theta)) + mu* (g*cos(theta))*d
now you can solve for mu ^^ Good luck.
0.091
7 answers