Question

a skier of mass 80 kg starts at the top of a 20m frictionless hill. after reaching the bottom he reaches a patch of snow 10m long with us =0.4.how fast is he going after he leaves the rough patch?

Answer 1

initial PE +W(by friction)=Final KE

So

mgh-us*mgd=0.5mv^2

So

80*9.8*20-0.4*80*9.8*10=0.5*80*v^2

So

v=17.71 m/s

Answer 2

speed at bottom of hill:

using conservation of energy:
mgh = .5mv²
v= 20m/s

friction force = us * normal force
= .4 * 80*10
= 320N

acceleration = -4m/s2 (- because this acceleration is in opposire direction of the velocity)

u = 20
a = -4
s= 10

using , v² = u² + 2*a*s

v = 17.89m/s