Understanding Velocity and Distance on an Inclined Plane with Friction

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SUMMARY

The discussion centers on calculating the distance a hockey player, Boom-Boom Slapshot, travels into snow after sliding down a 50m frictionless hill inclined at 35°. The correct acceleration on the incline is determined to be 5.62 m/s², leading to a maximum velocity of 23.71 m/s at the bottom. The distance traveled into the snow, accounting for a coefficient of kinetic friction of 0.50, is confirmed to be approximately 57m, contradicting an initial incorrect calculation of 175m. The confusion arises from the interpretation of the problem regarding the transition from the incline to the horizontal snow surface.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of kinematic equations
  • Familiarity with the concepts of friction and normal force
  • Basic trigonometry for calculating components of forces
NEXT STEPS
  • Study the application of Newton's second law in inclined planes
  • Learn about kinetic friction and its effects on motion
  • Explore the derivation and application of kinematic equations
  • Investigate the role of angles in force calculations on inclined surfaces
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to motion on inclined planes and friction.

nesan
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Homework Statement



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?

The Attempt at a Solution



I've figured out acceleration on the inclined plane which is 5.62m/s^2. But, I was having trouble with including friction into the other part as well as normal force.


a = sin35 * 9.8 = 5.62m/s^2

Using the formula Vf^2 = Vi^2 + 2(a)(d)

I found the velocity which is 23.71m/s

I got an answer of 175m but the textbook says the answer is 57m. The textbook has been wrong few times, I just want to confirm this, thank you. :)
 
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How did you use that velocity to get 175m?
When I calculated it, I got 57m.
 
Browne said:
How did you use that velocity to get 175m?
When I calculated it, I got 57m.

a1 = g*sin35° = 5.62 m/s²

Vmax = √[2*a1*s] = √[2*5.62*50] = 23.7 m/s

a2 = -g[sin35° - µ*cos35°] = -1.607 m/s²

d2 = -Vmax²/(2*a2) = 174.76 m

Did I do something wrong?
 
Ah, the problem is the interpretation of the question.
I think the problem was intended to mean that he starts traveling horizontally (at the bottom of the hill) when he enters the snow.
 
Browne said:
Ah, the problem is the interpretation of the question.
I think the problem was intended to mean that he starts traveling horizontally (at the bottom of the hill) when he enters the snow.

Can you expand on that please? I'm really confused after trying 4 times.

Thank you. Just pointing me in the right direction will do. :)
 
After he reaches Vmax, the incline becomes 0°
 
I need more help please, I'm kind of lost. =\
 
nesan said:
I need more help please, I'm kind of lost. =\
What is it that you don't understand regarding what Browne said in his last two posts?

It would be difficult to restate what he said in a manner that's any clearer !
 

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