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Old Dec 21st 2017, 04:27 AM   #1
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Question Urgent, please help

Hello, I'm new here. I've been trying to solve this problem in the image but I just can't seem to understand how you'd get numbers in the answer ( I know it is sqrt of 8gh/15 ) and it is supposedly done with energy methods.
I don't really know how to start
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Old Dec 21st 2017, 05:41 AM   #2
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This is strange. The attachment says "review" yet you don't even know how to start? That should worry you!

Start by assigning names to each of the forces. There is a downward force due to the weight of block B, call that "W". There is a downward force due to the weight of block A. We are told that blocks A and B are of equal mass so that is also "W". There are then three upward forces, the tensions in the ropes: $\displaystyle F_1$, the tension in the rope supporting A, $\displaystyle F_2$, the tension in the left rope supporting B, and $\displaystyle F_3$, the tension in the right rope supporting B. Since the first and two ropes are the same rope, separated only by a pulley, $\displaystyle F_1= F_2$.

The total force on A is $\displaystyle F_1- W$ and must be equal to the mass of A, W/g, times the acceleration of A. The total force on B is $\displaystyle F_2+ F_3= F_1+ F_3- W$ and must be equal to the mass of B, W/g, times the acceleration of B.
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Old Dec 21st 2017, 08:08 AM   #3
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I don`t understand this quite properly. I guess you may ask also at https://studydaddy.com/ they specialize in various cunning problems.
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Old Dec 21st 2017, 09:23 AM   #4
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Old Dec 21st 2017, 09:44 AM   #5
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when block B rises a distance $x$, block A falls a distance $2x$ and block B's speed will be half of block A's speed. Let the vertical separation of $3x=h$ and let the zero for gravitational potential energy be located at both blocks' initial rest position.

total final energy = total initial energy

$U_B + K_B + U_A + K_A = 0$

$mgx + \dfrac{1}{2}m\left(\dfrac{v}{2}\right)^2 - mg(2x) + \dfrac{1}{2}mv^2 = 0$

collect terms and simplify ...

$\dfrac{5v^2}{8} = gx$

multiply every term by $3$ ...

$\dfrac{15v^2}{8} = 3gx$

since $h = 3x$ ...

$\dfrac{15v^2}{8} = gh$

$v^2 = \dfrac{8gh}{15} \implies v = \sqrt{\dfrac{8gh}{15}}$

Last edited by skeeter; Dec 21st 2017 at 10:00 AM.
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