Posted by **leeann** on Sunday, May 27, 2012 at 3:58am.

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**, Sunday, May 27, 2012 at 5:45am
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**, Monday, May 28, 2012 at 12:13am
thank you. really appreciate it

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