Two blocks of masses m1 = 4.00 kg and m2 = 3.00 kg are placed in contact with each
other on a frictionless horizontal surface. A constant force F = 9.00 N is applied to the
block of mass m1. a) Determine the magnitude of the acceleration of the two-block
system. b) Determine the magnitude of the contact force, P12, between the two blocks.
(Answer a) 1.29 m/s2 b) 3.86 N)
Please show work if you want additional assistance
To find the magnitude of acceleration of the two-block system, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = m * a).
Let's consider the two blocks as a system. The force F is applied to the block with mass m1, and it will cause both blocks to accelerate together. However, since there is no friction, the acceleration of both blocks will be the same.
So, using Newton's second law for the two-block system, we have:
F = (m1 + m2) * a
Plugging in the given values:
9.00 N = (4.00 kg + 3.00 kg) * a
Now, we can solve for the acceleration (a):
a = 9.00 N / 7.00 kg
a = 1.29 m/s^2
Therefore, the magnitude of the acceleration of the two-block system is 1.29 m/s^2.
Now, let's find the magnitude of the contact force, P12, between the two blocks. Since there is no friction, the only forces at play are the applied force F and the contact force P12. The contact force acts in the opposite direction to the applied force.
According to Newton's third law of motion, the contact force between two objects is equal in magnitude but opposite in direction.
So, the magnitude of the contact force, P12, is 9.00 N (same as the applied force F), but in the opposite direction.
Therefore, the magnitude of the contact force between the two blocks is 9.00 N.