Assume that the three blocks in Figure P4.27 move on a frictionless surface and that a F = 46 N force acts as shown on the m = 2.5 kg block.

Figure P4.27.
(a) Determine the acceleration given this system.


m/s2 (to the right)
(b) Determine the tension in the cord connecting m and the 1.0 kg blocks.


N
(c) Determine the magnitude of the force exerted on the 2.0 kg block by the 1.0 kg block.
N

10

To find the acceleration of the system in Figure P4.27, we need to analyze the forces acting on each block.

(a) Determine the acceleration given this system:

1. First, let's calculate the net force acting on each block:
- The 2.5 kg block has a force F = 46 N acting on it.
- The 2.5 kg block is connected to the 1.0 kg block by a cord, and they move together. So, the 2.5 kg block also experiences the tension force from the cord.
- The 1.0 kg block is connected to the 2.5 kg block by a cord and experiences the tension force from the cord.
- There is no friction acting on any of the blocks.

2. Now, let's calculate the net force on each block in the x-direction:
- The 2.5 kg block experiences a force F = 46 N to the right.
- The 2.5 kg block and the 1.0 kg block are linked together by the cord, so they both have the same acceleration.
- The 1.0 kg block solely experiences the tension force.

3. Since the 2.5 kg and 1.0 kg blocks have the same acceleration, we can set up an equation using Newton's second law (F = ma) for both blocks:
- For the 2.5 kg block: F - T = m1a, where F is the applied force, T is the tension force, m1 is the mass of the 2.5 kg block, and a is the acceleration.
- For the 1.0 kg block: T = m2a, where m2 is the mass of the 1.0 kg block.

4. Let's substitute the given values into the equations:
- For the 2.5 kg block: 46 N - T = (2.5 kg) a
- For the 1.0 kg block: T = (1.0 kg) a

5. Now, we can solve these two equations simultaneously. Add the equations together to eliminate the T variable:
46 N - T + T = (2.5 kg + 1.0 kg) a
46 N = 3.5 kg a

6. Finally, solve for the acceleration:
a = (46 N) / (3.5 kg) ≈ 13.14 m/s^2 (to the right)

Therefore, the acceleration of the system is approximately 13.14 m/s^2 to the right.

(b) Determine the tension in the cord connecting the 2.5 kg and 1.0 kg blocks:

To find the tension force in the cord, we can substitute the acceleration value we found in the previous part into one of the equations we derived:

Using the equation for the 1.0 kg block: T = (1.0 kg) a
T = (1.0 kg) × (13.14 m/s^2)
T ≈ 13.14 N

Therefore, the tension in the cord connecting the 2.5 kg and 1.0 kg blocks is approximately 13.14 N.

(c) Determine the magnitude of the force exerted on the 2.0 kg block by the 1.0 kg block:

Since the 2.0 kg block is connected to the 1.0 kg block through the cord, the force exerted by the 1.0 kg block on the 2.0 kg block is equal to the tension force.

Therefore, the magnitude of the force exerted on the 2.0 kg block by the 1.0 kg block is approximately 13.14 N.