Question

The acceleration of block of mass 5 kg is: (neglect friction)

(A). \[5\text{ m}{{\text{s}}^{-2}}\] 

(B). \[10\text{ m}{{\text{s}}^{-2}}\]

(C). \[15\text{ m}{{\text{s}}^{-2}}\]

(D). \[20\text{ m}{{\text{s}}^{-2}}\] 

Answer 1

Hint: The figure has separate systems and different forces are acting on each system. Resolve the systems. Tension in the string and force applied are the main forces acting here. Analyse each force and determine external forces, if any and then make equations. Solve equations to find acceleration of block.

Formula used:

\[150\text{ - 2T = 0}\]

 \[\text{T = ma}\]

Complete step-by-step answer:

When there are net forces acting on a body then, the body is said to be in non-uniform motion that means it will accelerate.

When net forces acting on a body are 0 then, they are either in uniform motion or at rest.

We consider 2 separate systems in the figure and take the direction of force\[150\text{ N}\]as the positive direction.

Forces acting on the 10 kg block system are \[2\operatorname{T}\] and \[150\text{ N}\] where T is the tension in the string and \[150\text{ N}\] is being applied to it. 

Let us assume that a 10 kg block moving with acceleration a.

Therefore,

$150 - 2T = 10{a}$

$75 -T = 5 {a}$......(1)                                      

Force acting on the 5 kg block system is \[\text{T}\] and it is in the direction of\[150\text{ N}\] . 

Now the 5 kg block will also move with acceleration $2a$ because the second side of the pulley string is fixed.

Therefore,

\[\text{T = m2a}\]                                              - (2)

Where \[\text{m}\] is the mass of the block and \[\text{2a}\] is the acceleration of the block.

$T = 10a$

Adding  eq (1) and eq (2), we get, 

$75-T+T = 15{a}$

\[\therefore \text{ a = 5 m}{{\text{s}}^{-2}}\] 

Therefore the acceleration of 5kg block comes out to be \[\text{10 m}{{\text{s}}^{-2}}\] 

So, the correct answer is “Option  B”.

Note: Tension \[\text{T}\] acting on 5kg block is in the direction of force \[150\text{ N}\] while \[\text{2T}\] acting on \[10\text{ kg}\] block is in the opposite direction. No external forces are acting on the system, all given forces are considered internal forces. The motion of the pulley will be the same as the motion of a 10 kg block.