- #1

gungo

- 27

- 1

## 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?

(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

v2^2=v1^2+2ad

Fnet=ma

Ff=μkFn

## The Attempt at a Solution

a) FN=9.8(0.170)=1.666

Ff=0.005(1.6660)=0.00833

Fnet=ma

0.00833=0.170a =0.049

v2^2=v1^2+2ad

v2^2=(21.2)^2+2(0.049)(58.5)

v2=21.3, but the answer is 21.1

b)Ff=0.047(1.666)

Ff=0.078302

0.078302=0.170a=0.4606

v2^2=v1^2+2ad

I'll use the correct v2

21.1^2=21.2^2+2(0.4606)d

-4.23=0.9212d

d=-4.59 but the answer is 6.24

Not sure where I'm going wrong?