Question

A hockey puck is sliding across a frozen pond with an initial speed of 9.5 m/s. It comes to rest after sliding a distance of 37.4 m. What is the coefficient of kinetic friction between the puck and the ice

Answer 1

Answer:

The coefficient of kinetic friction between the puck and the ice is $\mu _{k} =$ 0.12

Explanation:

Given :

Initial speed  $v_{o} = 9.5 \frac{m}{s}$

Displacement $x = 37.4$ m

From the kinematics equation,

  $v^{2} - v^{2} _{o} = 2ax$

Where $v^{2} =$ final velocity, in our example it is zero ($v =0$), $a =$ acceleration.

   $a =- \frac{90.25}{ 2 \times 37.4}$

   $a =- 1.21 \frac{m}{s^{2} }$

From the formula of friction,

  $F =- \mu _{k } N$

Minus sign represent friction is oppose the motion

Where $N = mg$ ( normal reaction force )

 $ma = -\mu _{k} m g$                                                  ( ∵ $g = 9.8 \frac{m}{s^{2} }$ )

So coefficient of friction,

 $\mu_{k} = \frac{1.21}{9.8}$

 $\mu_{k} = 0.12$

Therefore, the coefficient of kinetic friction between the puck and the ice is  $\mu _{k} =$ 0.12 .