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## Homework Statement

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 traveling 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?

## Homework Equations

Fnet=ma

Fg=mg

Fk=μk.Fn

Vf[itex]^{2}[/itex]=Vi[itex]^{2}[/itex]+2ad

## The Attempt at a Solution

a.) Fnet=Fk

ma=μk.Fn

ma=μk.mg

a=[itex]\frac{(0.05)(0.170kg)(9.8m/s^{2})}{0.170kg}[/itex]

=0.049m/s[itex]^{2}[/itex]

Vf[itex]^{2}[/itex]=Vi[itex]^{2}[/itex]+2ad

=[itex]\sqrt{(21.2)^{2}-(0.049)(58.5)}[/itex]

=21.13m/s

b)Fnet=Fk

ma=μk.Fn

ma=μk.mg

a=[itex]\frac{(0.047)(0.170kg)(9.8m/s^{2})}{0.170kg}[/itex]

=0.04606m/s[itex]^{2}[/itex]

.. then i don't know what's next..