Posted by Jake on Sunday, December 20, 2009 at 5:52pm.
for some reason I keep on getting 1.2 seconds for (b) and .86 m for (a)
The back of the book has the same answers for (a) but has 1.5 seconds for (b) i do not see what I'm doing wrong.
Can someone confirm that I'm wrong or right???
A blcok is given an intial speed of 3.0 m/s up a 22 degree plane. (a) How far up the plane will it go? (b) How mcuh time elapses before it returns to its starting point? Assume coefficent of kinetic frction is .17

Physics  bobpursley, Sunday, December 20, 2009 at 6:09pm
If it goes up .86m,then
average velocityup=1.5m/s
then on the way down,
finalKE=initialPEworkfriction.
1/2 mv^2=mg .86Sin.86mu*mg*Cos22
v^2=2g(.86sin22.17*Cos22)
vfinal=1.80
avg v down=.90
avg v total= distance/time= .86*2/(.86/1.5 + .86/.9)=1.12m/s
distance:.86*2; avg velocity 1.12m/s
time= 1.53 seconds

Physics  drwls, Sunday, December 20, 2009 at 6:14pm
(a) The block will proceed uphill until friction work, 0.17 M g X cos 22, plus potential energy gain M g sin 22 equals initial kinetic energy.
V^2/2 = 0.17*9.81*0.927X + 9.81*0.375 = (1.546 +3.679)X
X = V^2/(2*5.225) = 0.86 m
(b) The average speed going up is 1.5 m/s, so it takes 0.86/1.5 = 0.573 s to go up. Coming down, the final speed is given by
Vf^2/2 = g X sin 22  g*cos 22*0.17*X
= 3.16  1.33 = 1.83 m^2/s^2
Vf = 1.91 m/s
Vav(down) = 0.957 m/s
Time spent going back down = X/Vav= 0.899 s
Total elapsed time = 0.573 + 0.899 = 1.47 s Call it 1.5 s since we have only two significant figures.
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