If 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
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
2 answers
(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.
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.
3 answers