Constrained Motion - NEET Physics Questions

Question 1:

The force F needed to keep the block at equilibrium in given figure is (pulley and string are massless)

neet constrained motion questions

1. Mg/5

2. Mg/4

3. Mg/2

4. Mg/3

View Answer

For Equilibrium,

2F = Mg ⇒ F = Mg/2

Question 2:

If acceleration of block m1 is a downward then acceleration of block m2 will be:

neet constraint motion questions

1. 2a upward

2. a upward

3. a/2 upward

4. 2a downward

View Answer

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:

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

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

Question 3:

What is the minimum value of F needed so that block begins to move upward on frictionless incline plane as shown?

nlm constrainted motion quesiton 9

1. Mg tan(θ/2)

2. Mg cot(θ/2)

3. Mg.sinθ/(1+sinθ)

4. Mgsin(θ/2)

View Answer

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 =

$Mg \sin \theta$

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

Now use the identity:

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

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