Question

Plz answer this question

Answer 1

The force applied on 3 kg is more than the force applied on 2 kg

=> acceleration (a) will be in forward direction.


Now 3 kg will apply a force on 2 kg.
However, this force will not be equal to 10kg as 3kg will resist some force.

Let's say R is the force 3 kg applies on 2kg.


So when we will draw FBDs (refer in attachment) we will find that 2 kg will also apply a force R on 3 kg according to Newton's third law.

This means 3 kg has two forces acting on it,

10N and R. (refer the attachment)

As we said above that acceleration is in forward direction,
=> 10N > R

So net force F = 10N - R

Now We know that F = ma

=> 10N - R = 3a ..........(equation 1)

Similarly on block 2kg,

Since the acceleration is forward,

=> R > 5N (refer the attachment)

=> Net force = R - 5N

again using F = ma

=> R - 5N = 2a (equation 2)

Adding equation 2 and equation 1,

10N - R + R - 5N = 3a + 2a

=> 5 Newton = 5a

=> a = 5/5

=> a = 1 m/s²

Now acceleration will be the same for both blocks as they are in contact.

So from equation 2,

R - 5 = 2a

=> R - 5 = 2(1)

=> R - 5 = 2

=> R = 2 + 5

=> R = 7N


We know that R is the force that acts in between the blocks

So Your answer = R = 7 Newton.



Hope you understand