A skier starts from rest at the top of a frictionless incline of height 20.0 m,

Question

A skier starts from rest at the top of a frictionless incline of height 20.0 m, as shown in the figure. At the bottom of the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is 0.188. Neglect air resistance.

A skier starts from rest at the top of a frictionless incline of height 20.0 m, as shown in the figure. At the bottom of the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is 0.188. Neglect air resistance. (a) Find the skier's speed at the bottom. m/s (b) How far does the skier travel on the horizontal surface before coming to rest? m Find the horizontal distance the skier travels before coming to rest if the incline also has a coefficient of kinetic friction equal to 0.188. Assume that ? = 20.0?. m

Explanation / Answer

a)mgh = 1/2 * mv2

or v2=2gh = 2*10*20 = 400

v=20 m/s

b) frictional force = mg = 0.188 * m * 10

Deacceleration = -0.188*10=-1.88

u=20,v=0,a=-1.88,S=?

S= (v2-u2)/2a = -400/-3.76 = 106.38 m

c)Work done by friction on incline = mgcos * h/cos = 0.188 * m * 10 * 20 = 37.6 m

So Energy at bottom = mgh - 37.6m = 162.4m

For velocity at bottom

1/2 * mv2 =162.4 m

or v2=324.8

v=18 m/s

frictional force on horizontal track = mg = 0.188 * m * 10

Deacceleration = -0.188*10=-1.88

u=18,v=0,a=-1.88,S=?

S= (v2-u2)/2a = -400/-3.76 = 86.17 metre

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