what is the value for the coefficient of kinetic friction ?

Question

A hockey puck is hit on a frozen lake and starts moving with a speed of 13.8 m/s. Five seconds later, its speed is 5.10 m/s.

a.) What is its average acceleration ?
b.) What is the average value of the coefficient of kinetic friction between puck and ice?
c.) How far does the puck travel during the 5.00 s interval?

Answer 1

1) a = kg

where a = acceleration
k is the coefficient of kinetic friction
G = the gravitational acceleration = 9.8ms-1

“a” can be calculated by the formula

Vf = V0 + at (1)

5.1m/s = 13.8m/s + a 5s
5.1- 13.8 = 5a
-8.7 = 5a
Hence a = 8.7/5 = -1.74ms-1

2) k = V02/2gx (2)

Where x is the distance traveled by te puck during 5s at an average deceleration of -1.74ms-1
We use the equation Vf2 –V02 = 2ax
5.12 – 13.82 = (2)(-1.74) (x) =47.3m
Hence x = 47.3m

Plug this value in equation (2) to get the value of kinetic friction

k = V02/2gx

k = 13.82/2 * 9.8* 47.3

Note: I am not able to get the equation symbols right! so you might be confused. If you want the full answer in MS Word format I can send it to your email as attachment.
k = 0.21

Answer 2

arbiter00...'s help could grant you with the acceleration. For the coefficient of kinetic friction, u, the rigidity that brought about that acceleration is given through F = m*a. The rigidity is brought about through friction it quite is given through u*N = u*m*g. Set both expressions equivalent to one yet another - via the indisputable fact that's a similar rigidity.

Answer 3

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.

Answer 4

velocity = initial velocity + acceleration*time

The equation and the rest of the equations should already be in your textbook or notes.

Answer 5

http://en.wikipedia.org/wiki/Friction