Physics · Laws Of Motion · NEET
No. The applied force pushes the whole system (both blocks). The contact force only has to accelerate the SECOND block. So contact force = (second block's mass) x a, which is always less than the applied force. In the NEET 2024 problem the applied force is 10 N but the contact force is only 6 N.
The force you apply must move the total mass (m1 + m2). But the block at the front (the one being pushed) has less mass than the total, so it needs a smaller force to reach the same acceleration. Contact force = m2 x a, and m2 is only part of the total, so this force is smaller than the applied force.
Step 1: Treat both blocks as one system and find common acceleration a = F / (m1 + m2). Step 2: Isolate the block being pushed (draw its free body diagram). The only horizontal force on it is the contact force N. Apply Newton's second law: N = m2 x a. That single number is the contact force.
Yes, the contact force value changes but the method is the same. The common acceleration a = F / (m1 + m2) stays the same because total mass and applied force are the same. But now the contact force must accelerate the OTHER block, so use that block's mass. Push a 3 kg block against a 2 kg block with 10 N: a = 2 m/s^2, contact force = 2 x 2 = 4 N.
Always the mass of the block that the contact force is acting ON, that is, the block being pushed (the one NOT receiving the external force directly). The external force acts on the back block; the back block pushes the front block; so use the front block's mass.
A horizontal force of 10 N is applied to block A, which is in contact with and pushes block B on a frictionless horizontal surface. The masses of A and B are 2 kg and 3 kg respectively. The force exerted by block A on block B is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the normal push that one block gives the other where they touch. By Newton's third law, block A pushes B forward with force N, and B pushes A back with the same N. Its value is N = (mass of pushed block) x common acceleration.
First a = F / (m1 + m2). Then the contact force on the front block of mass m2 is N = m2 x a = F x m2 / (m1 + m2). The force is proportional to the pushed block's share of the total mass.
Because the two blocks are in contact and move together, they must have the same acceleration. Finding it from the whole system first is the fastest way, then you plug it into one block's equation to get the contact force.
Here yes. The horizontal contact force between the two touching faces is a normal (perpendicular) push, so it is a normal reaction force acting horizontally between the block faces.
Then each block also feels friction. You still find the common acceleration first, but now include friction forces in the system equation, then isolate one block including its friction to solve for the contact force.
