Question

62. Two blocks of masses 2 kg and 1 kg are in contact with each other on a frictionless table. When a horizontal force of 3.0 N is applied to the block of mass 1 kg the value of the force of contact between the two blocks is - (1) 2 N (2) 3 N (3) 5 N (4) 1 N​

Answer 1

Question

Two blocks of masses 2 kg and 1 kg are in contact with each other on a frictionless table. When a horizontal force of 3.0 N is applied to the block of mass 1 kg the value of the force of contact between the two blocks is -

(1) 2 N

(2) 3 N

(3) 5 N

(4) 1 N​

Answer:

As we know force= mass * acceleration

                      acceleration =force/mass

                          a=3/2+1

                          a=1m/sec2

Now force is applied horizontly on 1kg mass so

force on 1kg block = 1 *1=1 N

force on 2kg block =2*1=2N

so force of contact =2N-1N=1N

hence option 4

itz mridu

Answer 2

Answer:

(1)2N

Step1:-

Using sys boundry

$a = \frac{fnet}{mtotal}$

a=3/3

=1ms-2

Step2:-

Now take any of the block and using Fnet=ma we will calulate contact force (N')

●USING 2KG BLOCK

N'=2×1=2 OR

●IF WE USE 1KG BLOCK THEN

3-N'=1×1

3-1=N'

N'=2Newton