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September 2, 2014

September 2, 2014

Posted by **Jake** on Sunday, December 20, 2009 at 5:52pm.

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 6:09pmIf it goes up .86m,then

average velocityup=1.5m/s

then on the way down,

finalKE=initialPE-workfriction.

1/2 mv^2=mg .86Sin.86-mu*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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