Two blocks are pushed along a horizontal frictionless surface by a force of 20 newtons to the right. The force acts on the 3 kg block, which in turn pushes the 2kg block sitting next to it. The force that the 2-kilogram block exerts on the 3-kilogram block is......
8N left
8 newtons
8N to the left
Well, it seems like these blocks are having some serious trust issues. The 3-kilogram block might want to take a step back and have a serious talk with the 2-kilogram block about boundaries. Alright, let's get back to business.
Since there is no friction to worry about, the force exerted by the 2-kilogram block on the 3-kilogram block is equal in magnitude but opposite in direction to the force exerted by the 3-kilogram block on the 2-kilogram block. So, it's also 20 newtons but going in the opposite direction. It's like a never-ending tug of war, but with blocks!
To find the force that the 2-kilogram block exerts on the 3-kilogram block, we can use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
Since the 2-kilogram block is pushed by the 3-kilogram block, the force exerted on the 3-kilogram block is equal in magnitude but opposite in direction to the force exerted on the 2-kilogram block.
Given that the applied force is 20 newtons to the right, the force exerted by the 2-kilogram block on the 3-kilogram block will also be 20 newtons but to the left. This is because the reaction force is always equal in magnitude but opposite in direction to the action force.
So, the force that the 2-kilogram block exerts on the 3-kilogram block is 20 newtons to the left.
figure acceleartion first.
a= 20/5= 4m/s^2
So to make the 2kg block accelerate at 4m/s^2...
F=ma, and of course that is the force between the 2,3 blocks