How Long Does It Take for a Skier to Stop on a Bunny Hill?

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The discussion revolves around calculating the stopping time and distance for a skier on a 5-degree bunny hill with a mass of 25kg and an initial speed of 3.5 m/s, factoring in a coefficient of kinetic friction of 0.20. Initial calculations for gravitational force and friction are presented, leading to the determination of acceleration as 1.95 m/s². The user initially misinterprets the forces acting on the skier, mistakenly considering friction as the only force. Clarification is provided that both gravitational and frictional forces must be considered to accurately calculate net force and subsequent motion. The conversation emphasizes the importance of correctly identifying all forces in physics problems to arrive at accurate solutions.
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A 25kg skier on a "bunny" hill that is inclined at 5 degrees has an intial velocity of 3.5 m/s. The coefficient of kinetic friction is 0.20

a) calculate the time taken for the skier to come to a stop
b) the distance traveled down the hill..

what i have so far is..

ive calculated is
Fg =mg = 25*(9.8) = 245N
Fn = mgcos 5 = 25*(9.8) * cos 5 = 244.07N
Friction = (mu) Fn = 0.20* 244.07 = 48.81N
F=ma thus 48.81 = 25a thus a= 1.95 m/s^2

a = Vf - Vi / delta T ... delta T = 1.79s

am i right so far?
 
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Check your net force equation. According to yours, friction is the only force acting on the skier. This would mean that he is being pushed by friction and friction only.
 
ok so i have force normal, force of gravity and fictional force acting on the skier, so how do i add up those three?
 
You add up the ones parallel to the direction of acceleration. The normal force is not relevant, but the other component that makes up mg is relevant.
 
so what i have now is Fg + Fk - mg sin 5 = net force
is what i have now correct?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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