NEET Constrained Motion MCQs & PYQs

Net Force acting on the block in upward direction is 4T which will balance weight of the object. So, 

⇒ 4T= 800 N 

⇒T = 200 N

Writing equations of motion for the objects as,

40 - T = 4 × 2a     ---(i)

and 2T -40 = 4a    -----(ii)

Solving (i) and (ii)

a= 2 m/s²

So, acceleration of object A is 4 m/s² and for object B is 2 m/s²

As F= 340 N. Force acting on object A in upward direction is F/2 i.e 170 N. This force is insufficient to pull the object upwards. (weight of A is greater than 170 N)

So, a1= 0 m/s²

Force acting on B in upward direction is 170 N and weight is 10g or 100 N. Net force in upward direction is 70 N. so,

a2= 7 m/s²

Let us assume tension force in the wire is T. Then net force acting on man and crate in upward direction is 2T. Gravitational force acting in downward direction is (50+100) g = 1500 N. Writing equation of motion for the system.

2T- 1500 = 150 × a

⇒ 2T- 1500 = 150 × 5

⇒ 2T = 750 +1500 = 2250

⇒ T = 1125 N

When two objects are connected by a string component of velocity along the string remains equal.

v₁ cosθ = v ₀

⇒v₁= v₀/ cosθ

For Equilibrium,

2F = Mg ⇒ F = Mg/2

In the movable pulley system, tension in the string connecting m2 and m3 is:

$$T = \frac{2 m_2 m_3 g}{m_2 + m_3}$$

Since this tension acts twice to balance

$m_1$

, we equate:

$$2T = m_1 g \Rightarrow \frac{4 m_2 m_3 g}{m_2 + m_3} = m_1 g$$

Cancelling

$g$

and rearranging gives:

$$\boxed{ \frac{4}{m_{1}} = \frac{1}{m_{2}} + \frac{1}{m_{3}} }$$

For ideal pulleys, the product of tension and acceleration remains constant:

$$T_1 a = T_2 a_2$$

Since

$T_2 = 2T_1$

, substitute to get:

$$T_1 a = 2T_1 a_2 \Rightarrow a_2 = \frac{a}{2}$$

$\boxed{\text{So, } M_2 \text{ accelerates upward with } \frac{a}{2}}$

🔹 Step-by-step:

• Block of mass M is on a frictionless incline of angle
  $\theta$ 
• Force
  $F$ 

  is applied via a pulley system, split into two components:

  • One acts up along the incline: F
  • One acts horizontally, which when resolved along the incline becomes: F
    $F \cos \theta$ 

Total upward force along incline =

$F + F \cos \theta$

Downward component of weight =

$Mg \sin \theta$

🔹 For the block to just start moving upward:

$$F+F\cos \theta =Mg\sin \theta$$ $$F(1+\cos \theta )=Mg\sin \theta$$ $$F = \frac{Mg \sin \theta}{1 + \cos \theta}$$

Now use the identity:

$$\frac{\sin \theta}{1 + \cos \theta} = \cot\left(\frac{\theta}{2}\right)$$

 Final Answer:

$$\boxed{F = Mg \cot\left(\frac{\theta}{2}\right)}$$