Sunday
March 26, 2017

Post a New Question

Posted by on .

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?

  • physics - ,

    Use equation mgh + mgh(initial) = -1/2mv^2 - 1/2 mv(initial) + W

  • physics - ,

    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.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question