two blocks of masses m1 and m2 (m1>m2) are placed on a frictionless table in contact with each other. A horizontal force of magnitude F is applied to the block of mass m1. if P is the magnitude of the contact force between the blocks, what are the net forces acting on m1 and m2? Assume that "to the right" is the positive direction
a) on m1: F+P on m2: -P
b) on m1: F-P on m2: +P
c) on m1: -P on m2: F+P
d) on m1: +P on m2: F-P
e) on m1: +P on m2: F+P
(b)
m1•a=F-P
m2•a=P
The correct answer is:
a) on m1: F+P, on m2: -P
Explanation:
When a horizontal force F is applied to the block of mass m1, the block will experience a force in the direction of the applied force (to the right). This force, F, is the net force acting on m1.
Since the blocks are in contact with each other, they exert equal and opposite contact forces on each other. So, the block of mass m2 will experience a contact force in the opposite direction (to the left) with magnitude P. This contact force, -P, is the net force acting on m2.
Therefore, the correct statement is that the net force acting on m1 is F+P, and the net force acting on m2 is -P.
To find the answer, we need to analyze the forces acting on each block individually.
Let's start with block m1. The only external force acting on m1 is the applied horizontal force F. Since "to the right" is considered the positive direction, the force F is positive.
Now, let's consider block m2. The only force acting on m2 is the contact force P between the two blocks. Since the contact force acts in the opposite direction to the applied force, it is negative.
Therefore, the answer is:
On m1: F+P
On m2: -P
So, the correct option is (a) on m1: F+P on m2: -P.