Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Dec 21st 2017, 03:27 AM #1 Junior Member   Join Date: Dec 2017 Posts: 1 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 Attached Thumbnails
 Dec 21st 2017, 04:41 AM #2 Senior Member   Join Date: Aug 2010 Posts: 388 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. daftpower likes this.
 Dec 21st 2017, 07:08 AM #3 Banned   Join Date: Dec 2017 Posts: 1 I don`t understand this quite properly. I guess you may ask also at https://studydaddy.com/ they specialize in various cunning problems.
 Dec 21st 2017, 08:23 AM #4 Senior Member   Join Date: Aug 2010 Posts: 388 Please do not link to pay sites here.
 Dec 21st 2017, 08:44 AM #5 Senior Member     Join Date: Aug 2008 Posts: 113 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 09:00 AM.

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