A hockey puck is hit on a frozen lake and starts moving with a speed of 21 m/s. Exactly 5.00 s later, its speed is 4.9 m/s. Its acceleration is -3.22m/s^2. It travels 64.75m during this 5.00 s interval.
What is the average value of the coefficient of kinetic friction between puck and ice?
2007-11-14 09:20:44 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Coefficient of Kinetic Friction= -a/g
I'm not quite sure about the average value part because the coefficient of kinetic friction should remain constant between any two surface areas depending on the material. In this
case a=-3.22 and g=-9.8 and the rest of the numbers are not needed because you are given "a", which usually is asked to be solved for using the rest of the values, but not in this case.
Coefficient of Kinetic Friction =(3.22/9.8) = .32857
2007-11-14 10:00:46 · answer #1 · answered by buckeyesean27 2 · 0⤊ 0⤋
arbiter00...'s help should give you the acceleration. For the coefficient of kinetic friction, u, the force that caused that acceleration is given by F = m*a. The force is caused by friction which is given by u*N = u*m*g. Set the 2 expressions equal to each other - because it's the same force.
2016-05-23 04:27:29 · answer #2 · answered by krystle 3 · 0⤊ 0⤋