Question
Work n power
A ski starts from rest and slides down a 22° incline 110 mlong. Use energy methods.
(a) If the coefficient of friction is 0.09, what is the ski'sspeed at the base of the incline?
m/s
(b) If the snow is level at the foot of the incline and has thesame coefficient of friction, how far will the ski travel along thelevel?
mwhat equations do i use?
Answers
a)the friction force Fr = -mgcos(22)*0.09Δh=110m*sin(22)=41.21mthe kenetic energy at the bottom of the incline is:KE=potential energy - work done by friction=>(1/2)mv2=mgΔh-mgcos(22)*0.09*110mv=√[2gΔh-gcos(22)*0.09*110m]=26.79m/sb)now the friction force is: F=mg*0.09F=-mg*0.09=maa=-g*0.09=-0.882m/s2when it stop vf = 0, let the distance be s, wehave:vf2 - v2 = 2ass =-v2/2a=-(26.79m/s)2/(-2*0.882m/s2)= 406.86 m..
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