Question

The system shown in Figure P8.11 consists of a light, inextensible cord, light, frictionless pulleys, and blocks of equal mass. Notice that block B is attached to one of the pulleys. The system is initially held at rest so that the blocks are at the same height above the ground. The blocks are then released. Find the speed of block A at the moment the vertical separation of the blocks is.

Short answer

The speed of block A at the moment the vertical separation of the blocks is $\sqrt{\frac{8g{h}^{}}{15}}$

Step-by-step solution

Step by step solution:Step 1: Given data

Two blocks of equal mass are attached in a pulley system.

Determine potential and kinetic energy

The potential energy of a mass at a height from the ground is

$P=mgh$

Here, g is the acceleration due to gravity.

The kinetic energy of a mass moving with a velocity v is

$k=\frac{1}{2}m{v}^2$

Determine the speed of block A

Since block B is attached to the pulley that is on the other end of the rope attached to block A, if block B moves up by distance x, block B moves down by distance 2x. The velocity of block A is also twice of the velocity of block B. To make the separation between the two blocks h, block B has to move up By $\frac{h}{3}$and thus block A moves down by $\frac{2h}{3}$. In an isolated system, the sum of change in kinetic energy and potential energy is zero. Let the velocity of block A be and the mass of the blocks be m. Then, from equations (I) and (II)

Thus, the required velocity is $\frac{\sqrt{8gh}}{\sqrt{15}}$