Question

Given system is released from rest.Time taken by 2kg to reach the pulley is sqrt(5)x sec .Find the value of x .(3kg come to rest just after hiting the ground)​

Answer 1

Explanation:

For the 2kg block

T=ma=2a

For the 3 kg block

3g-T=3a

3g-2a=3a

3g=5a

a= 3g/5

now the 3kg block comes to rest after travelling 5m

and the for the 2kg block block

u=0

v= ?

t= _/5 x

s= 10m

v²=u²+2as

v²=2as

v²= 2×10×6

v= 2_/30

now v=u+at

2_/30=_/5×x×6

2_/6=6x

_/6=3x

x= (3/2)^1/2

Answer 2

Given:

The system shown in the figure

Distance covered by the 2kg block = 10m

Time taken by the 2kg block to reach the pulley = √5 x seconds

To Find:

The value of x

Solution:

• For finding the value of x, first, we need to make a Free Body Diagram(FBD) of the system.
• An FBD shows the direction and magnitude of all the forces acting on all the objects in a system.
• Let 'a' be the acceleration of the system.
• We get the following:

1. Force 3mg acting downwards on the 3kg block.
2. Tension T in the string attached to the 3kg block acting upwards.
3. Force 2mg acting downwards on the 2kg block, balanced by the Normal reaction of the plane.
4. Tension T in the string attached to the 2kg block acting away from it.

Now, we will follow these steps:

1. Find an expression for the acceleration of the 2kg block.

• As we know, Force = mass × acceleration
• As the plane is smooth, no friction is acting on the block.
• So, we get:

                                  2a = T                         (1)

                             

• Similarly for the 3kg block:

                       3a = 3g - T

                   ⇒ 3a = 3g - 2a                       (from (1) )

                   ⇒ 5a = 3g

                   ⇒ a = $\frac{3g}{5}$                                  (2)

2. Find an expression for the time taken by the 2kg block to reach the pulley.

• Since we know the acceleration and distance to be covered by the block, we can use the following formula:

                              s = ut + $\frac{1}{2}$at²

         where s is the displacement of the body

                    u is the initial velocity

                    t is the time taken

                    a is the acceleration

• We know that initially, the 2kg block was at rest, hence u = 0.
• So, the formula gets simplified to:

                                 s = $\frac{1}{2}$at²

                            ⇒ 10 = $\frac{1}{2}$ × $\frac{3g}{5}$ t²

                           ⇒ 10 = $\frac{3}{10}$ × 10 × t²

                           ⇒ t² = $\frac{10}{3}$            

                           ⇒ t = $\sqrt{\frac{10}{3} }$

3. Equate expression of time derived with the given value.

• The given value of time is √5 x.
• So, we get:

                             √5 x = $\sqrt{\frac{10}{3} }$

                            Squaring both sides

                           ⇒ 5 $x^{2}$ = $\frac{10}{3}$

                           ⇒ x² = $\frac{2}{3}$

                           ⇒ x = $\sqrt{\frac{2}{3} }$

Hence the value of x is $\sqrt{\frac{2}{3} }$.