How Far Will Boom-Boom Slapshot Travel in Snow After Sliding Down an Icy Hill?

[Brodo17]

Boom-Boom Slapshot, Canadian hockey star, slides down a 50m long ice covered hill on his skates. The frictionless hill is inclined at 35° to the horizontal. Once he reaches the bottom of the hill, the ice is covered with deep snow that has a coefficient of kinetic friction of .50
How far into the snow will BoomBoom go before coming to rest?



Equations
acceleration = fparallel / mass
vf^2 = 2a Δd
f = µFN
FN = opposite of Fg which is mass x 9.8


Attempt
I drew out the problem and came to the conclusion that since I was working with an inclined plane that the mass of the hockey player didnt matter. I figured out (at least I am pretty sure) that the acceleration is Sin35 * 9.8 = 5.62m/s
Now I just don't know how to figure out the negative acceleration when he hits the snow. Because in order to calculate force friction, I need to know his mass. If someone could help me out it would be much appreaciated! Thank you

 

[tiny-tim]

Welcome to PF!

Hi Brodo17! Welcome to PF!

Boom-Boom Slapshot, Canadian hockey star, slides down a 50m long ice covered hill on his skates. The frictionless hill is inclined at 35° to the horizontal. Once he reaches the bottom of the hill, the ice is covered with deep snow that has a coefficient of kinetic friction of .50
How far into the snow will BoomBoom go before coming to rest?

Forget acceleration … the angle is only needed for calculating the height …

this is an energy problem …

use conservation of energy on the hill,

and the work-energy theorem ( work done = energy lost ) on the flat

 

[lanedance]

Hi Brodo17

tinytim is right on the money... just thought I would add your acceleration looks correct & you could do it that way, but the looking at the energy will save you a heap of time & effort

 

[Brodo17]

Thanks guys!