Two blocks of 2 kg and 1 kg are in contact on a frictionless table. If a force of 3 N is applied on 2 kg block, then the force of contact between the two blocks will be :

Question

Two blocks of 2 kg and 1 kg are in contact on a frictionless table. If a force of 3 N is applied on 2 kg block, then the force of contact between the two blocks will be : (A) 0 N (B) 1 N (C) 2 N (D) 3 N

Tardigrade Admin · Accepted Answer

Acceleration of both the blocks taken together =(3/[2+1])=1 m / s 2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/6f963eaa0a120177ff4c95a027a4d372-.png" /> Hence, the contact force between the two blocks is equal to 1 kg × 1 m / s 2=1 N

Q. Two blocks of $2 k g$ and $1 k g$ are in contact on a frictionless table. If a force of $3 N$ is applied on $2 k g$ block, then the force of contact between the two blocks will be :

0 N

1 N

2 N

3 N

Acceleration of both the blocks taken together
$= \frac{3}{[ 2 + 1 ]} = 1 m / s^{2}$

Hence, the contact force between the two blocks is equal to
$1 k g \times 1 m / s^{2} = 1 N$

Q. Two blocks of 2kg and 1kg are in contact on a frictionless table. If a force of 3N is applied on 2kg block, then the force of contact between the two blocks will be :

0 N

1 N

2 N

3 N

Acceleration of both the blocks taken together =[2+1]3​=1m/s2 Hence, the contact force between the two blocks is equal to 1kg×1m/s2=1N

Q. Two blocks of $2 k g$ and $1 k g$ are in contact on a frictionless table. If a force of $3 N$ is applied on $2 k g$ block, then the force of contact between the two blocks will be :

Acceleration of both the blocks taken together
$= \frac{3}{[ 2 + 1 ]} = 1 m / s^{2}$

Hence, the contact force between the two blocks is equal to
$1 k g \times 1 m / s^{2} = 1 N$