A sledder starting from rest, slides down a 10-m-high hill At the bottom of the

Question

A sledder starting from rest, slides down a 10-m-high hill At the bottom of the hill is a long hortzontal patch of rough snow. The hill is nearty frictioniess, but the coeficient of kinetic friction between the sled and the rough snow at the bottom is unknown. The sled slides along the horizontal rough patch for 50 m. What is the coefficient of kinetic friction on the rough snow? Hints: apply work-energy theorem. Before: y = 10m v, = 0 m/s After: yi Frictionless v, = 0 m/s vOm/s Ar = 50 m 0.50 O b.0.20 Oc 0.30 , 0.25 Oe.0.40

Explanation / Answer

Motion from top of hill to Bottom of hill :

h = height = 10 m

Vi = initial velocity at the top = 0 m/s

Vf = final velocity at the bottom of hill

Using conservation of energy :

Total energy at Top = Total energy at bottom

(0.5) m Vi2 + mgh = (0.5) m Vf2

Vi2 + 2gh = Vf2

02 + 2 x 9.8 x 10 = Vf2

Vf = 14 m/s

Motion on rough patch ::

a = retardation on the patch

d = displacement = 50 m

Vf = final velocity = 0     since skater comes to rest finally

Vi = speed the bottom of hill = 14 m/s

Using the equation

Vf2 = Vi2 + 2 a d

02 = 142 + 2 a (50)

a = 1.96 m/s2

the retardation due to frictional force on a horizontal surface is given as

a = uk g                     uk = Coefficient of kinetic friction

1.96 = uk (9.8)

uk = 0.2

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