Question

A hockey puck on a frozen pond with an initial speed of 13.7 m/s stops after sliding a distance of 216.9 m. Calculate the average value of the coefficient of kinetic friction between the puck and the ice.

Answer 1

Answer:

0.04

Explanation:

From

$ma=\mu mg$

Here m is mass, a is acceleration, $\mu$  is the coefficient of kinetic friction and g is acceleration due to gravity.

Making $\mu$ the subject of the formula

$\mu=\frac {a}{g}$

From kinematics, we know that

$v^{2}-u^{2}=2as$

Here, v is final speed, u is initial speed, a is acceleration and s is distance moved.

Making a the subject of the formula then

$a=\frac {v^{2}-u^{2}}{2s}$

Since it stops, the final velocity is 0 while the initial speed is given as 13.7 m/s

Substituting 0 for v, 13.7 for u, 216.9 m for s then

$a=\frac {0^{2}-13.7^{2}}{2\times 216.9}=-0.432664822 m/s^{2}$

Taking g as $9.81 m/s^{2}$then using the formula

$\mu=\frac {0.432664822}{9.81}=0.044104467\approx 0.04$