Question

A 110 ghockey puck sent sliding over ice is stopped in 15 mby the frictional force on it from the ice.

(a) If its initial speed is 6.0 m/s, what is the magnitude of the frictional force?

(b) What is the coefficient of friction between the puck and the ice?

Short answer

$$\begin{gathered} a)f=0.13N \\ b){\mu }_K=0.12 \end{gathered}$$

Step-by-step solution

Given data

• The mass of the hockey puck is $m=110g$.
• The distance traveled before stopping is$s=15\mathrm{m}$.
• The initial velocity of the hockey puck is$v_0=6\mathrm{m}/\mathrm{s}$.
• The final velocity of the hockey puck is $v=0\mathrm{m}/\mathrm{s}$.

Understanding the concept

This problem is based on the kinematic equations of motion in which the motion of an object is described at constant acceleration. Use kinematics equations to calculate the acceleration and frictional force. We can calculate the kinetic friction coefficient from the frictional force.

Force:

$v^2=v_o^2+2as$

(a)Calculate the magnitude of the frictional force if its initial speed is 6.0 m/s

The free body diagram of the hockey puck is shown below:

To calculate the acceleration expression, use kinematics equations:

$v^2=v_o^2+2as$

The final velocity is zero, and the expression can be written as:

$a=-v_o^2/2s$

To calculate frictional force, use Newton’s second law of motion:

$$\begin{gathered} f=ma \\ =\frac{m{v}_o^2}{2s} \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} f=\frac{\left(0.115\mathrm{kg}\right)\times \left(6\mathrm{m}/\mathrm{s}\right)}{2\times \left(15\mathrm{m}\right)} \\ f=0.13\mathrm{N} \end{gathered}$$

Thus, the magnitude of the friction force is$0.13\mathrm{N}$.

(b) Calculate the coefficient of friction between the puck and the ice

To calculate the kinetic frictional coefficient, use the formula for kinetic friction:

${\mu }_{k}mg=f$

Substitute the values in the above expression, and we get,

$$\begin{gathered} {\mu }_k=\frac{0.13\mathrm{N}}{\left(0.115\mathrm{kg}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)} \\ {\mu }_k=0.12 \end{gathered}$$

Thus, the coefficient of friction is 0.12.