Posted by angel on Monday, August 1, 2011 at 2:16am.
Sin¤= 1/3
¤=19.47'
Fnet= Fg// - Ff
Fnet= mg.sin¤ -Ff
=2(9.8)sin19.47-0.35
Fnet= 6.183N
a= Fnet/m
a= 6.183/2
a=3.091m/s^2
(Vf^2)=(Vi^2) +2as
Vf^2= (0)+2(3.091)(3)
Vf^2= 18.546
Vf= 4.31m/s downward.
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a skier starts from rest and slide 9 m down a slope in 3 s. at what time, after starting, will the skier acquire a velocity of 24m/s? assume constant acceleration.
how can you solve it using work power and energy?
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