Question

Consider the situation shown in the figure. The system is released from rest and the block A travels a distance 5m in downward direction take g = 10m/s)

Answer 1

Answer:

$the answer is1ms$

Answer 2

Given,

A rope pulley system with body A having a mass of 10kg

Body B of mass 5kg.

To find,

Work done by the tension of Block B.

Solution,

Looking at the rope pulley system,

If block A is pulled down by x distance, block B will move up by x/2 distance.

Thus, when the acceleration of block A is, $a$ acceleration of block B will be $\frac{a}{2}$

For Block A,

       Weight acting downwards - tension in the upward direction = 10*a

      $100N - T = 10a$ -----------eq(1)

Similarly, for block B, we have

The tension acting on block B will be 2T.

Tension in the upward direction - Weight acting downwards = 5*a/2

     $2T - 50N = \frac{5}{2}a$ --------------eq(2)

Solving, eq(1) and eq(2)

Value of tension (T) = $\frac{100}{3}$N

Tension acting on block B = 2T = $\frac{200}{3}N$

Work done by tension on block B= Focrce*displacement

                                                        $= 2T*2.5m\\ = 166.667J$

Therefore, work done by the tension on block B is $166.667J.$