Posted by leeann on .
A 16kg sled starts up a 28 degree incline with a speed of 2.4 m/s . The coefficient of kinetic friction is = 0.27.
part a)How far up the incline does the sled travel?
part b)What condition must you put on the coefficient of static friction if the sled is not to get stuck at the point determined in part (a)? i got 0.53.
part c)If the sled slides back down, what is its speed when it returns to its starting point
can you please do me a favor and write out the work and the final answer. for some reason when i do it it comes out with the wrong final answer.

physics please respond 
Damon,
ke at bottom = (1/2) m v^2 = .5*2.4^2 * m
= 2.88 m
work done going distance x up ramp
= mu m g cos 28 x = .27*9.8*cos 28 * m
= 2.34 m x
Pe at top = m g h = m*9.8*x *sin 28
= 4.60 m x
so
4.60 x = 2.88  2.34 x
x = .415 meters (note mass does not matter)
b)
m g sin 28 = mu m g cos 28
tan 28 = mu
mu = .532 agree
c) work done = twice work going up
=2*2.34 m x = 4.68 m x = 4.68*.415 m
= 1.94 m
so
Ke = initial Ke  work done by friction
(1/2) m v^2 = 2.88 m  1.94 m
v^2 = 1.88
v = 1.37 m/s 
physics please respond 
leeann,
thank you. really appreciate it