Atwood's machine height problem

[tony873004]

Figure 8-20 http://www.webassign.net/walker/08-20b_alt.gif

14. [Walker2 8.P.025.] For the Atwood's machine shown in Figure 8-20, suppose that m2 has an initial upward speed of v = 0.17 m/s.

How high does m2 rise above its initial position before momentarily coming to rest, given that m1 = 3.3 kg and m2 = 4.1 kg?

Here's what I did:

F_{1} = T - m_{1}g = m_{1}a
F_{2} = m_{2}g - T = m_{2}a

and since the tensions must equal each other

(m_{2} - m_{1})g = (m_{1} + m_{2})a
or
a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} g

a = \frac{4.1 - 3.3}{3.3 + 4.1} 9.81

a = 1.06054

I'm pretty confident that the answer is correct so far because the book uses this exact method in their example. Only the masses and velocity are changed. But the book's example stops at acceleration. Our problem asks for a distance, so...

d = \frac{\Delta V^2}{2a}

d = \frac{0.17^2}{2*1.06054}

d = 0.0136 m

d = 1.36 cm

But Webassign, the online homework system says WRONG

But the same problem in the book, using m_{1}=3.7 , m_{2} = 4.1 , v = 0.2 gives an answer of 4cm, which is exactly what I get using the same method. How can this method be good using the book's numbers, but wrong using the numbers in the online problem? Could the online grading be wrong? Wouldn't be the first time

 

[puregoodboi]

hmm not sure

 

[bomerman218]

Its much easier if you use work energy to solve this problem. Also the tensions are not equal- only in the static case but the relative position, velocity, and acceleration are.