Question

A 75­kg snowboarder has an initial velocity of 5.0 m/s at the top of a 28° incline. After sliding down the 110­m long incline (on which the coefficient of kinetic friction is μk=0.18), the snowboarder has attained a velocity v. The snowboarder then slides along a flat surface (on which μk=0.15) and comes to a rest after a distance x.

a. Use energy calculations to determine x.

b. Use Newton’s 2nd Law to find the snowboarder’s acceleration while on the incline

and on the flat surface. Then use these accelerations to find x.

Answer 1

m = 75 kg , u =5 m/s , μk=0.18, heta =28degrees

L =110 m

sin(28) = h/L

h = Lsin(28)

Work done by friction force is equal to change in mechanical energy

Wf = Kf+Uf - Ki -Ui

- fk.L = (1/2)mv^2 + 0 - (1/2)mu^2 -mgh

fk =μkN = μkmgcos(28)

-μkmgcos(28)L = (1/2)mv^2 - (1/2)mu^2 -mgLsin(28)

v^2 = u^2 +2gLsin(28) -2μkgLcos(28)

v^2 = (5*5) +(2*9.8*110*sin(28)) - (2* 0.18*9.8*100*cos(28)) =694.52

v= 26.354

Work done by fricition on flat surface is

Wf = Kf -Ki

-μkmg.x = 0 -(1/2)mv^2

x = v^2/2μkg

x = (694.52)/(2*0.15*9.8)

x =236.23 m

(b) From Newton second law along vertical direction

Fy =0

N = mgcos(28)

From Newton second law along horizontal direction

Fx =ma

-fk +mgsin(28) =ma

fk =μkN =μkmgcos(28)

-μkmgcos(28) +mgsin(28) =ma

a = -μkgcos(28) +gsin(28)

a = -(0.18*9.8*cos(28))+(9.8*sin(28))

a = 3.0433m/s^2

From kinematic equations

v^2-u^2 =2as

v^2 = (5*5) +(2*3.0433*110)

v = 26.345 m

From kinematic equation for flate surface

0 -v^2 =2μkg*x

x = v^2/2μkg* = (26.345*26.345)/(2*0.15*9.8)

x = 236.23m