Question

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?
my solution:
Fnet=Fk
ma=μk.Fn
ma=μk.mg
a=(0.05)(0.170kg)(9.8m/s2)0.170kg
=0.049m/s2

Vf2=Vi2+2ad
=(21.2)2−(0.049)(58.5)−−−−−−−−−−−−−−−−−−√
=21.13m/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?
my solution:
Fnet=Fk
ma=μk.Fn
ma=μk.mg
a=(0.047)(0.170kg)(9.8m/s2)0.170kg
=0.04606m/s2

.. then i don't know what's next..
i got part a right but i'm having trouble with part b... please help.

Answer 1

ma=F(fr) =μN=μmg
a= μg.
s=(v₀²-v²)/2a=
=(v₀²-v²)/2μg =>
v=sqrt{v₀²-2μgs}=
=sqrt{21.2²-2•0.005•9.8•58.5} =
=21.06 m/s.

For the 2nd case:
s=(v₀²-v²)/2μ₁g=
=(21.2²-21.06²)/2•0.047•9.8=6.42 m

Answer 2

Thank you

Answer 3

Bruuh i dont understand

Answer 4

I, Rafae Adil, am an idiot

Answer 5

You bafoon, you used 0.5 as mu in ur calculations but its really 0.005 in the question

Answer 6

you dumbo, you didn't put a slash and it looks like you multiply

Answer 7

How did you get F net is equal to F k