Question

practice

1. a 0.170 kg hockey puck is initially moving at 21.2 m/s w along the ice. the coefficient of kinetic friction for the puck and the ice is 0.005.

(a) what is the speed of the puck after travelling 58.5 m? ans: 21.1 m/s
(b) after being played on for a while, the ice becomes rougher and the coefficient of kinetic friction increases to 0.047. how far will the puck travel if its initial and final speeds are the same as before? ans: 6.24 m

Explanation:

Step1: Calculate the acceleration

The frictional force \(F_f=\mu_kmg\). According to Newton's second law \(F = ma\), so \(a=-\mu_kg\).
For part (a), \(\mu_k = 0.005\), \(g = 9.8m/s^2\), then \(a=- 0.005\times9.8=-0.049m/s^2\)
We use the kinematic equation \(v^{2}=v_{0}^{2}+2ax\). Given \(v_{0} = 21.2m/s\), \(x = 58.5m\), \(a=-0.049m/s^2\)
Substitute into the equation: \(v^{2}=(21.2)^{2}+2\times(- 0.049)\times58.5\)
\(v^{2}=449.44-5.733\)
\(v^{2}=443.707\)
\(v=\sqrt{443.707}\approx21.1m/s\)

Step2: Calculate the distance for part (b)

For part (b), \(\mu_k = 0.047\), then \(a=-0.047\times9.8=-0.4606m/s^2\)
Using the kinematic equation \(v^{2}=v_{0}^{2}+2ax\), and we know \(v = 21.1m/s\), \(v_{0} = 21.2m/s\)
Rearrange the equation for \(x\): \(x=\frac{v^{2}-v_{0}^{2}}{2a}\)
Substitute \(v = 21.1m/s\), \(v_{0} = 21.2m/s\), \(a=-0.4606m/s^2\)
\(x=\frac{(21.1)^{2}-(21.2)^{2}}{2\times(-0.4606)}=\frac{445.21 - 449.44}{-0.9212}=\frac{-4.23}{-0.9212}\approx6.24m\)

Answer:

(a) \(21.1m/s\)
(b) \(6.24m\)