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.
can some one please work it out and explain how to get the answer i really need help with it
α=total force/ total mass
=9N/(4kg + 3kg)
=1.29 m/s^2
Sure! Let's solve this problem step-by-step.
a) To determine the magnitude of the acceleration of the two-block system, we need to use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
The net force acting on the system is the applied force (F) minus the force of friction. Since the surface is frictionless, the force of friction is zero.
So, the net force on the system is equal to the applied force (F) only.
Net force (F_net) = F = 9.00 N
Now, we can apply Newton's second law:
F_net = (m1 + m2) * a
Substituting the given values:
9.00 N = (4.00 kg + 3.00 kg) * a
Simplifying the equation:
9.00 N = 7.00 kg * a
Dividing both sides by 7.00 kg:
a = 9.00 N / 7.00 kg = 1.29 m/s^2
Therefore, the magnitude of the acceleration of the two-block system is 1.29 m/s^2.
b) To determine the magnitude of the contact force (P12) between the two blocks, we need to consider the forces acting on each block.
For block m1, it experiences the applied force (F) and the contact force (P12) between the two blocks.
For block m2, it experiences the contact force (P12) only.
The contact force (P12) between the two blocks is equal in magnitude but opposite in direction for each block.
Using Newton's second law for both blocks:
For block m1:
F_net1 = m1 * a
F - P12 = m1 * a
For block m2:
F_net2 = m2 * a
P12 = m2 * a
Substituting the given values and the previously calculated acceleration:
For block m1:
9.00 N - P12 = 4.00 kg * 1.29 m/s^2
9.00 N - P12 = 5.16 N
For block m2:
P12 = 3.00 kg * 1.29 m/s^2
P12 = 3.87 N
Therefore, the magnitude of the contact force (P12) between the two blocks is 3.87 N.
I hope this helps! Let me know if you have any further questions.
To solve this problem, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration (F = ma).
a) To determine the magnitude of the acceleration of the two-block system, we need to consider the net force acting on the system. The applied force (9.00 N) acts on block m1, while block m2 experiences an equal and opposite force due to the interaction between the two blocks. Therefore, the net force acting on the system is 9.00 N.
Let's assume the acceleration of the two-block system as 'a'. We'll use this assumption to find the value of 'a'.
For block m1:
F = m1 * a
9.00 N = 4.00 kg * a
a = 9.00 N / 4.00 kg
a ≈ 2.25 m/s²
Therefore, the magnitude of the acceleration of the two-block system is approximately 2.25 m/s².
b) To determine the magnitude of the contact force (P12) between the two blocks, we can consider either block m1 or m2.
Using block m2:
F12 = m2 * a
F12 = 3.00 kg * 2.25 m/s²
F12 = 6.75 N
Therefore, the magnitude of the contact force, P12, between the two blocks is 6.75 N.